Method to test if a number is a perfect power?Detecting perfect squares faster than by extracting square rooteffective way to get the integer sequence A181392 from oeisA rarely mentioned fact about perfect powersHow many numbers such $n$ are there that $n<100,lfloorsqrtn rfloor mid n$Check perfect squareness by modulo division against multiple basesFor what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.Do there exist any positive integers $n$ such that $lfloore^nrfloor$ is a perfect power? What is the probability that one exists?finding perfect power factors of an integerProve that the sequence contains a perfect square for any natural number $m $ in the domain of $f$ .Counting Perfect Powers

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Method to test if a number is a perfect power?


Detecting perfect squares faster than by extracting square rooteffective way to get the integer sequence A181392 from oeisA rarely mentioned fact about perfect powersHow many numbers such $n$ are there that $n<100,lfloorsqrtn rfloor mid n$Check perfect squareness by modulo division against multiple basesFor what pair of integers $(a,b)$ is $3^a + 7^b$ a perfect square.Do there exist any positive integers $n$ such that $lfloore^nrfloor$ is a perfect power? What is the probability that one exists?finding perfect power factors of an integerProve that the sequence contains a perfect square for any natural number $m $ in the domain of $f$ .Counting Perfect Powers













4












$begingroup$


Is there a general method for testing numbers to see if they are perfect $n$th powers?



For example, suppose that I did not know that $121$ was a perfect square. A naive test in a code might be to see if
$$lfloorsqrt121rfloor=sqrt121$$



But I imagine there are much more efficient ways of doing this (if I'm working with numbers with many digits).










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
    $endgroup$
    – Alex R.
    2 hours ago










  • $begingroup$
    Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
    $endgroup$
    – Servaes
    2 hours ago










  • $begingroup$
    @Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
    $endgroup$
    – D.B.
    1 hour ago










  • $begingroup$
    Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
    $endgroup$
    – D.B.
    1 hour ago






  • 2




    $begingroup$
    @D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
    $endgroup$
    – Alex R.
    1 hour ago
















4












$begingroup$


Is there a general method for testing numbers to see if they are perfect $n$th powers?



For example, suppose that I did not know that $121$ was a perfect square. A naive test in a code might be to see if
$$lfloorsqrt121rfloor=sqrt121$$



But I imagine there are much more efficient ways of doing this (if I'm working with numbers with many digits).










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
    $endgroup$
    – Alex R.
    2 hours ago










  • $begingroup$
    Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
    $endgroup$
    – Servaes
    2 hours ago










  • $begingroup$
    @Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
    $endgroup$
    – D.B.
    1 hour ago










  • $begingroup$
    Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
    $endgroup$
    – D.B.
    1 hour ago






  • 2




    $begingroup$
    @D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
    $endgroup$
    – Alex R.
    1 hour ago














4












4








4


2



$begingroup$


Is there a general method for testing numbers to see if they are perfect $n$th powers?



For example, suppose that I did not know that $121$ was a perfect square. A naive test in a code might be to see if
$$lfloorsqrt121rfloor=sqrt121$$



But I imagine there are much more efficient ways of doing this (if I'm working with numbers with many digits).










share|cite|improve this question











$endgroup$




Is there a general method for testing numbers to see if they are perfect $n$th powers?



For example, suppose that I did not know that $121$ was a perfect square. A naive test in a code might be to see if
$$lfloorsqrt121rfloor=sqrt121$$



But I imagine there are much more efficient ways of doing this (if I'm working with numbers with many digits).







number-theory perfect-powers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Chase Ryan Taylor

4,45021531




4,45021531










asked 2 hours ago









D.B.D.B.

1,29518




1,29518







  • 1




    $begingroup$
    One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
    $endgroup$
    – Alex R.
    2 hours ago










  • $begingroup$
    Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
    $endgroup$
    – Servaes
    2 hours ago










  • $begingroup$
    @Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
    $endgroup$
    – D.B.
    1 hour ago










  • $begingroup$
    Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
    $endgroup$
    – D.B.
    1 hour ago






  • 2




    $begingroup$
    @D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
    $endgroup$
    – Alex R.
    1 hour ago













  • 1




    $begingroup$
    One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
    $endgroup$
    – Alex R.
    2 hours ago










  • $begingroup$
    Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
    $endgroup$
    – Servaes
    2 hours ago










  • $begingroup$
    @Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
    $endgroup$
    – D.B.
    1 hour ago










  • $begingroup$
    Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
    $endgroup$
    – D.B.
    1 hour ago






  • 2




    $begingroup$
    @D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
    $endgroup$
    – Alex R.
    1 hour ago








1




1




$begingroup$
One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
$endgroup$
– Alex R.
2 hours ago




$begingroup$
One very cheap, necessary condition is that $x^2pmod 4equiv 0,1$.
$endgroup$
– Alex R.
2 hours ago












$begingroup$
Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
$endgroup$
– Servaes
2 hours ago




$begingroup$
Are you given numbers $k$ and $n$ and asked to check whether $k$ is an $n$-th power? Or are you given just $k$ and asked to check whether $k$ is a perfect power?
$endgroup$
– Servaes
2 hours ago












$begingroup$
@Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
$endgroup$
– D.B.
1 hour ago




$begingroup$
@Servaes, I was considering the first case, where I know both k and n and trying to see if $k = a^n,$ a a positive integer.
$endgroup$
– D.B.
1 hour ago












$begingroup$
Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
$endgroup$
– D.B.
1 hour ago




$begingroup$
Wait, @Alex R. Looking at your first comment, what about $x^2 = 40 = 0 (mod 4)$. Yet, $40$ is not a perfect square.
$endgroup$
– D.B.
1 hour ago




2




2




$begingroup$
@D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
$endgroup$
– Alex R.
1 hour ago





$begingroup$
@D.B.: Hence it's a necessary condition: if $x^2$ is a perfect square, then $x^2equiv 0,1pmod4$. The other direction gives: if $yequiv 2,3pmod4$, then $y$ cannot be a perfect square.
$endgroup$
– Alex R.
1 hour ago











5 Answers
5






active

oldest

votes


















6












$begingroup$

See Detecting perfect powers in essentially linear time - Daniel J. Bernstein:



https://cr.yp.to/papers/powers-ams.pdf






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    In the specific case where you already know not only the number being checked but also the power, as the question's comment by the OP to Servaes states, then you have something like



    $$k = a^n tag1labeleq1$$



    where $k$ and $n$ are known integers, but with $a$ being an unknown value to check whether or not it's an integer. In this case, taking natural logarithms of both sides (you could use any base, but I suspect that implementation wise $e$ will likely at least be the fastest one, if not also the most accurate) gives



    $$ln(k) = nln(a) ; Rightarrow ; ln(a) = fracln(k)n ; Rightarrow ; a = e^fracln(k)n tag2labeleq2$$



    On a computer, this will give a floating point value that would be, even for large values of $k$, relatively close to the correct value of $a$.



    You can now use any number of algorithms to relatively quickly & easily determine $a$ if it's an integer, or show it's not an integer. For example, you can start with the integer part obtained in eqrefeq2, call it $a_1$, to determine $k_1$. If $k_1$ is not correct, then if it's less than $k$, check $a_2 = a_1 + 1$, else check $a_2 = a_1 - 1$, and call the new result $k_2$. If $k_2$ is still not correct, add or subtract the integer amount (making sure it's at least 1) of $left|frack -k_2k_1 - k_2right|$ to $a_2$ to get a new $a_1$ value to check. Then repeat these steps as many times as needed. In almost all cases, I believe it should take very loops to find the correct value. However, note you should also include checks in case there is no such integer $a$, with this usually being seen when one integer value gives a lower result & the next higher gives a higher result (or higher result & next lower integer gives a lower result).






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
      $endgroup$
      – miracle173
      19 mins ago


















    0












    $begingroup$

    My suggestion on a computer is to run a root finder.



    Given a value $y$, one way is to hard-code the first couple and then use an integer-valued binary search starting with $y/2$, which is logarithmic in $y$ and thus linear (since input takes $ln y$.



    You can also write down the Newton's method recurrence and see if it converges to an integer or not, should become clear after the first couple of steps, once the error becomes small enough.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      I don't think it's linear, given that you need to square the proposed number at every split.
      $endgroup$
      – Alex R.
      1 hour ago



















    0












    $begingroup$

    There are many powerful codes that factorize a number to its prime factors in a non-polynomial time (for more information you can refer to Integer Factorization on Wikipedia) . Once an integer was factorized as follows$$n=p_1^alpha_1times p_2^alpha_2timescdots times p_m^alpha_m$$then by defining $d=gcd(alpha_1,alpha_2,cdots ,alpha_m)$ we can say that $n$ is a full $d$-th power.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
      $endgroup$
      – gt6989b
      2 hours ago






    • 2




      $begingroup$
      For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
      $endgroup$
      – Arturo Magidin
      2 hours ago










    • $begingroup$
      This is orders-of-magnitude slower than just computing the square-root even by classical methods.
      $endgroup$
      – Alex R.
      2 hours ago










    • $begingroup$
      I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
      $endgroup$
      – MPW
      2 hours ago










    • $begingroup$
      I mean the most efficient that is possible so far
      $endgroup$
      – Mostafa Ayaz
      2 hours ago


















    0












    $begingroup$

    It is at least possible to do this in polynomial time. Assume $n$ is a $k$-bit number and you want to find positive integers $a$ and $b$ such that $$a^b=ntag1$$ or prove that such numbers don't exists.



    We have $$n<2^k$$ because $n$ is a $k$-bit number and so $$blt k$$



    We can simply check for all possible $b$ if there is an $a$ such that $(1)$ holds. For given $b$ we can try to find $a$ by bisection. This bisection checks $O(log n)=O(k)$ different $a$. A check is the calculation of $a^b$. This can be achieved by multiplying powers of $a$ by $a$. These powers of $a$ are smaller than $n$. So we multiply $k$-bit numbers at most $b(lt k)$ times. A multiplication of two $k$-bit numbers needs $O(k^2)$ time. So all in all the algorithm needs $O(k^2)$ multiplications o $k$-bit numbers, which means $O(k^4)$ time.






    share|cite|improve this answer









    $endgroup$












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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      See Detecting perfect powers in essentially linear time - Daniel J. Bernstein:



      https://cr.yp.to/papers/powers-ams.pdf






      share|cite|improve this answer









      $endgroup$

















        6












        $begingroup$

        See Detecting perfect powers in essentially linear time - Daniel J. Bernstein:



        https://cr.yp.to/papers/powers-ams.pdf






        share|cite|improve this answer









        $endgroup$















          6












          6








          6





          $begingroup$

          See Detecting perfect powers in essentially linear time - Daniel J. Bernstein:



          https://cr.yp.to/papers/powers-ams.pdf






          share|cite|improve this answer









          $endgroup$



          See Detecting perfect powers in essentially linear time - Daniel J. Bernstein:



          https://cr.yp.to/papers/powers-ams.pdf







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Alex J BestAlex J Best

          2,32611226




          2,32611226





















              1












              $begingroup$

              In the specific case where you already know not only the number being checked but also the power, as the question's comment by the OP to Servaes states, then you have something like



              $$k = a^n tag1labeleq1$$



              where $k$ and $n$ are known integers, but with $a$ being an unknown value to check whether or not it's an integer. In this case, taking natural logarithms of both sides (you could use any base, but I suspect that implementation wise $e$ will likely at least be the fastest one, if not also the most accurate) gives



              $$ln(k) = nln(a) ; Rightarrow ; ln(a) = fracln(k)n ; Rightarrow ; a = e^fracln(k)n tag2labeleq2$$



              On a computer, this will give a floating point value that would be, even for large values of $k$, relatively close to the correct value of $a$.



              You can now use any number of algorithms to relatively quickly & easily determine $a$ if it's an integer, or show it's not an integer. For example, you can start with the integer part obtained in eqrefeq2, call it $a_1$, to determine $k_1$. If $k_1$ is not correct, then if it's less than $k$, check $a_2 = a_1 + 1$, else check $a_2 = a_1 - 1$, and call the new result $k_2$. If $k_2$ is still not correct, add or subtract the integer amount (making sure it's at least 1) of $left|frack -k_2k_1 - k_2right|$ to $a_2$ to get a new $a_1$ value to check. Then repeat these steps as many times as needed. In almost all cases, I believe it should take very loops to find the correct value. However, note you should also include checks in case there is no such integer $a$, with this usually being seen when one integer value gives a lower result & the next higher gives a higher result (or higher result & next lower integer gives a lower result).






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
                $endgroup$
                – miracle173
                19 mins ago















              1












              $begingroup$

              In the specific case where you already know not only the number being checked but also the power, as the question's comment by the OP to Servaes states, then you have something like



              $$k = a^n tag1labeleq1$$



              where $k$ and $n$ are known integers, but with $a$ being an unknown value to check whether or not it's an integer. In this case, taking natural logarithms of both sides (you could use any base, but I suspect that implementation wise $e$ will likely at least be the fastest one, if not also the most accurate) gives



              $$ln(k) = nln(a) ; Rightarrow ; ln(a) = fracln(k)n ; Rightarrow ; a = e^fracln(k)n tag2labeleq2$$



              On a computer, this will give a floating point value that would be, even for large values of $k$, relatively close to the correct value of $a$.



              You can now use any number of algorithms to relatively quickly & easily determine $a$ if it's an integer, or show it's not an integer. For example, you can start with the integer part obtained in eqrefeq2, call it $a_1$, to determine $k_1$. If $k_1$ is not correct, then if it's less than $k$, check $a_2 = a_1 + 1$, else check $a_2 = a_1 - 1$, and call the new result $k_2$. If $k_2$ is still not correct, add or subtract the integer amount (making sure it's at least 1) of $left|frack -k_2k_1 - k_2right|$ to $a_2$ to get a new $a_1$ value to check. Then repeat these steps as many times as needed. In almost all cases, I believe it should take very loops to find the correct value. However, note you should also include checks in case there is no such integer $a$, with this usually being seen when one integer value gives a lower result & the next higher gives a higher result (or higher result & next lower integer gives a lower result).






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
                $endgroup$
                – miracle173
                19 mins ago













              1












              1








              1





              $begingroup$

              In the specific case where you already know not only the number being checked but also the power, as the question's comment by the OP to Servaes states, then you have something like



              $$k = a^n tag1labeleq1$$



              where $k$ and $n$ are known integers, but with $a$ being an unknown value to check whether or not it's an integer. In this case, taking natural logarithms of both sides (you could use any base, but I suspect that implementation wise $e$ will likely at least be the fastest one, if not also the most accurate) gives



              $$ln(k) = nln(a) ; Rightarrow ; ln(a) = fracln(k)n ; Rightarrow ; a = e^fracln(k)n tag2labeleq2$$



              On a computer, this will give a floating point value that would be, even for large values of $k$, relatively close to the correct value of $a$.



              You can now use any number of algorithms to relatively quickly & easily determine $a$ if it's an integer, or show it's not an integer. For example, you can start with the integer part obtained in eqrefeq2, call it $a_1$, to determine $k_1$. If $k_1$ is not correct, then if it's less than $k$, check $a_2 = a_1 + 1$, else check $a_2 = a_1 - 1$, and call the new result $k_2$. If $k_2$ is still not correct, add or subtract the integer amount (making sure it's at least 1) of $left|frack -k_2k_1 - k_2right|$ to $a_2$ to get a new $a_1$ value to check. Then repeat these steps as many times as needed. In almost all cases, I believe it should take very loops to find the correct value. However, note you should also include checks in case there is no such integer $a$, with this usually being seen when one integer value gives a lower result & the next higher gives a higher result (or higher result & next lower integer gives a lower result).






              share|cite|improve this answer











              $endgroup$



              In the specific case where you already know not only the number being checked but also the power, as the question's comment by the OP to Servaes states, then you have something like



              $$k = a^n tag1labeleq1$$



              where $k$ and $n$ are known integers, but with $a$ being an unknown value to check whether or not it's an integer. In this case, taking natural logarithms of both sides (you could use any base, but I suspect that implementation wise $e$ will likely at least be the fastest one, if not also the most accurate) gives



              $$ln(k) = nln(a) ; Rightarrow ; ln(a) = fracln(k)n ; Rightarrow ; a = e^fracln(k)n tag2labeleq2$$



              On a computer, this will give a floating point value that would be, even for large values of $k$, relatively close to the correct value of $a$.



              You can now use any number of algorithms to relatively quickly & easily determine $a$ if it's an integer, or show it's not an integer. For example, you can start with the integer part obtained in eqrefeq2, call it $a_1$, to determine $k_1$. If $k_1$ is not correct, then if it's less than $k$, check $a_2 = a_1 + 1$, else check $a_2 = a_1 - 1$, and call the new result $k_2$. If $k_2$ is still not correct, add or subtract the integer amount (making sure it's at least 1) of $left|frack -k_2k_1 - k_2right|$ to $a_2$ to get a new $a_1$ value to check. Then repeat these steps as many times as needed. In almost all cases, I believe it should take very loops to find the correct value. However, note you should also include checks in case there is no such integer $a$, with this usually being seen when one integer value gives a lower result & the next higher gives a higher result (or higher result & next lower integer gives a lower result).







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 49 mins ago

























              answered 56 mins ago









              John OmielanJohn Omielan

              4,2162215




              4,2162215











              • $begingroup$
                you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
                $endgroup$
                – miracle173
                19 mins ago
















              • $begingroup$
                you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
                $endgroup$
                – miracle173
                19 mins ago















              $begingroup$
              you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
              $endgroup$
              – miracle173
              19 mins ago




              $begingroup$
              you skip important steps of your algorithm. How do you calculate $a = e^fracln(k)n$. What is the time and space complexity of this calculation? How big is the difference of the exact value of $e^fracln(k)n$ and the calculated value of $e^fracln(k)n$? Without calculating all this bounds it is not possible to decide if the algorithm is efficient.
              $endgroup$
              – miracle173
              19 mins ago











              0












              $begingroup$

              My suggestion on a computer is to run a root finder.



              Given a value $y$, one way is to hard-code the first couple and then use an integer-valued binary search starting with $y/2$, which is logarithmic in $y$ and thus linear (since input takes $ln y$.



              You can also write down the Newton's method recurrence and see if it converges to an integer or not, should become clear after the first couple of steps, once the error becomes small enough.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I don't think it's linear, given that you need to square the proposed number at every split.
                $endgroup$
                – Alex R.
                1 hour ago
















              0












              $begingroup$

              My suggestion on a computer is to run a root finder.



              Given a value $y$, one way is to hard-code the first couple and then use an integer-valued binary search starting with $y/2$, which is logarithmic in $y$ and thus linear (since input takes $ln y$.



              You can also write down the Newton's method recurrence and see if it converges to an integer or not, should become clear after the first couple of steps, once the error becomes small enough.






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I don't think it's linear, given that you need to square the proposed number at every split.
                $endgroup$
                – Alex R.
                1 hour ago














              0












              0








              0





              $begingroup$

              My suggestion on a computer is to run a root finder.



              Given a value $y$, one way is to hard-code the first couple and then use an integer-valued binary search starting with $y/2$, which is logarithmic in $y$ and thus linear (since input takes $ln y$.



              You can also write down the Newton's method recurrence and see if it converges to an integer or not, should become clear after the first couple of steps, once the error becomes small enough.






              share|cite|improve this answer









              $endgroup$



              My suggestion on a computer is to run a root finder.



              Given a value $y$, one way is to hard-code the first couple and then use an integer-valued binary search starting with $y/2$, which is logarithmic in $y$ and thus linear (since input takes $ln y$.



              You can also write down the Newton's method recurrence and see if it converges to an integer or not, should become clear after the first couple of steps, once the error becomes small enough.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 1 hour ago









              gt6989bgt6989b

              35.1k22557




              35.1k22557











              • $begingroup$
                I don't think it's linear, given that you need to square the proposed number at every split.
                $endgroup$
                – Alex R.
                1 hour ago

















              • $begingroup$
                I don't think it's linear, given that you need to square the proposed number at every split.
                $endgroup$
                – Alex R.
                1 hour ago
















              $begingroup$
              I don't think it's linear, given that you need to square the proposed number at every split.
              $endgroup$
              – Alex R.
              1 hour ago





              $begingroup$
              I don't think it's linear, given that you need to square the proposed number at every split.
              $endgroup$
              – Alex R.
              1 hour ago












              0












              $begingroup$

              There are many powerful codes that factorize a number to its prime factors in a non-polynomial time (for more information you can refer to Integer Factorization on Wikipedia) . Once an integer was factorized as follows$$n=p_1^alpha_1times p_2^alpha_2timescdots times p_m^alpha_m$$then by defining $d=gcd(alpha_1,alpha_2,cdots ,alpha_m)$ we can say that $n$ is a full $d$-th power.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
                $endgroup$
                – gt6989b
                2 hours ago






              • 2




                $begingroup$
                For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
                $endgroup$
                – Arturo Magidin
                2 hours ago










              • $begingroup$
                This is orders-of-magnitude slower than just computing the square-root even by classical methods.
                $endgroup$
                – Alex R.
                2 hours ago










              • $begingroup$
                I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
                $endgroup$
                – MPW
                2 hours ago










              • $begingroup$
                I mean the most efficient that is possible so far
                $endgroup$
                – Mostafa Ayaz
                2 hours ago















              0












              $begingroup$

              There are many powerful codes that factorize a number to its prime factors in a non-polynomial time (for more information you can refer to Integer Factorization on Wikipedia) . Once an integer was factorized as follows$$n=p_1^alpha_1times p_2^alpha_2timescdots times p_m^alpha_m$$then by defining $d=gcd(alpha_1,alpha_2,cdots ,alpha_m)$ we can say that $n$ is a full $d$-th power.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
                $endgroup$
                – gt6989b
                2 hours ago






              • 2




                $begingroup$
                For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
                $endgroup$
                – Arturo Magidin
                2 hours ago










              • $begingroup$
                This is orders-of-magnitude slower than just computing the square-root even by classical methods.
                $endgroup$
                – Alex R.
                2 hours ago










              • $begingroup$
                I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
                $endgroup$
                – MPW
                2 hours ago










              • $begingroup$
                I mean the most efficient that is possible so far
                $endgroup$
                – Mostafa Ayaz
                2 hours ago













              0












              0








              0





              $begingroup$

              There are many powerful codes that factorize a number to its prime factors in a non-polynomial time (for more information you can refer to Integer Factorization on Wikipedia) . Once an integer was factorized as follows$$n=p_1^alpha_1times p_2^alpha_2timescdots times p_m^alpha_m$$then by defining $d=gcd(alpha_1,alpha_2,cdots ,alpha_m)$ we can say that $n$ is a full $d$-th power.






              share|cite|improve this answer











              $endgroup$



              There are many powerful codes that factorize a number to its prime factors in a non-polynomial time (for more information you can refer to Integer Factorization on Wikipedia) . Once an integer was factorized as follows$$n=p_1^alpha_1times p_2^alpha_2timescdots times p_m^alpha_m$$then by defining $d=gcd(alpha_1,alpha_2,cdots ,alpha_m)$ we can say that $n$ is a full $d$-th power.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 48 mins ago

























              answered 2 hours ago









              Mostafa AyazMostafa Ayaz

              18.1k31040




              18.1k31040











              • $begingroup$
                not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
                $endgroup$
                – gt6989b
                2 hours ago






              • 2




                $begingroup$
                For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
                $endgroup$
                – Arturo Magidin
                2 hours ago










              • $begingroup$
                This is orders-of-magnitude slower than just computing the square-root even by classical methods.
                $endgroup$
                – Alex R.
                2 hours ago










              • $begingroup$
                I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
                $endgroup$
                – MPW
                2 hours ago










              • $begingroup$
                I mean the most efficient that is possible so far
                $endgroup$
                – Mostafa Ayaz
                2 hours ago
















              • $begingroup$
                not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
                $endgroup$
                – gt6989b
                2 hours ago






              • 2




                $begingroup$
                For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
                $endgroup$
                – Arturo Magidin
                2 hours ago










              • $begingroup$
                This is orders-of-magnitude slower than just computing the square-root even by classical methods.
                $endgroup$
                – Alex R.
                2 hours ago










              • $begingroup$
                I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
                $endgroup$
                – MPW
                2 hours ago










              • $begingroup$
                I mean the most efficient that is possible so far
                $endgroup$
                – Mostafa Ayaz
                2 hours ago















              $begingroup$
              not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
              $endgroup$
              – gt6989b
              2 hours ago




              $begingroup$
              not sure about efficient, I believe it's an NP-complete problem. But surely checking if it is a perfect square would be doable more efficiently that doing the full prime factorization?!
              $endgroup$
              – gt6989b
              2 hours ago




              2




              2




              $begingroup$
              For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
              $endgroup$
              – Arturo Magidin
              2 hours ago




              $begingroup$
              For algorithms, "efficiently and fast" usually means being both deterministic and polynomial time in the length of the input; there is no known polynomial time deterministic algorithm for factoring integers, so I would absolutely quibble with your use of "efficiently and fast".
              $endgroup$
              – Arturo Magidin
              2 hours ago












              $begingroup$
              This is orders-of-magnitude slower than just computing the square-root even by classical methods.
              $endgroup$
              – Alex R.
              2 hours ago




              $begingroup$
              This is orders-of-magnitude slower than just computing the square-root even by classical methods.
              $endgroup$
              – Alex R.
              2 hours ago












              $begingroup$
              I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
              $endgroup$
              – MPW
              2 hours ago




              $begingroup$
              I was going to suggest that if you factorize $n>1$ as you have shown, with all of the $p_i$ distinct and all of the $alpha_i>0$, then $n$ is a perfect power if and only if all of the $alpha_i$ are equal.
              $endgroup$
              – MPW
              2 hours ago












              $begingroup$
              I mean the most efficient that is possible so far
              $endgroup$
              – Mostafa Ayaz
              2 hours ago




              $begingroup$
              I mean the most efficient that is possible so far
              $endgroup$
              – Mostafa Ayaz
              2 hours ago











              0












              $begingroup$

              It is at least possible to do this in polynomial time. Assume $n$ is a $k$-bit number and you want to find positive integers $a$ and $b$ such that $$a^b=ntag1$$ or prove that such numbers don't exists.



              We have $$n<2^k$$ because $n$ is a $k$-bit number and so $$blt k$$



              We can simply check for all possible $b$ if there is an $a$ such that $(1)$ holds. For given $b$ we can try to find $a$ by bisection. This bisection checks $O(log n)=O(k)$ different $a$. A check is the calculation of $a^b$. This can be achieved by multiplying powers of $a$ by $a$. These powers of $a$ are smaller than $n$. So we multiply $k$-bit numbers at most $b(lt k)$ times. A multiplication of two $k$-bit numbers needs $O(k^2)$ time. So all in all the algorithm needs $O(k^2)$ multiplications o $k$-bit numbers, which means $O(k^4)$ time.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                It is at least possible to do this in polynomial time. Assume $n$ is a $k$-bit number and you want to find positive integers $a$ and $b$ such that $$a^b=ntag1$$ or prove that such numbers don't exists.



                We have $$n<2^k$$ because $n$ is a $k$-bit number and so $$blt k$$



                We can simply check for all possible $b$ if there is an $a$ such that $(1)$ holds. For given $b$ we can try to find $a$ by bisection. This bisection checks $O(log n)=O(k)$ different $a$. A check is the calculation of $a^b$. This can be achieved by multiplying powers of $a$ by $a$. These powers of $a$ are smaller than $n$. So we multiply $k$-bit numbers at most $b(lt k)$ times. A multiplication of two $k$-bit numbers needs $O(k^2)$ time. So all in all the algorithm needs $O(k^2)$ multiplications o $k$-bit numbers, which means $O(k^4)$ time.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  It is at least possible to do this in polynomial time. Assume $n$ is a $k$-bit number and you want to find positive integers $a$ and $b$ such that $$a^b=ntag1$$ or prove that such numbers don't exists.



                  We have $$n<2^k$$ because $n$ is a $k$-bit number and so $$blt k$$



                  We can simply check for all possible $b$ if there is an $a$ such that $(1)$ holds. For given $b$ we can try to find $a$ by bisection. This bisection checks $O(log n)=O(k)$ different $a$. A check is the calculation of $a^b$. This can be achieved by multiplying powers of $a$ by $a$. These powers of $a$ are smaller than $n$. So we multiply $k$-bit numbers at most $b(lt k)$ times. A multiplication of two $k$-bit numbers needs $O(k^2)$ time. So all in all the algorithm needs $O(k^2)$ multiplications o $k$-bit numbers, which means $O(k^4)$ time.






                  share|cite|improve this answer









                  $endgroup$



                  It is at least possible to do this in polynomial time. Assume $n$ is a $k$-bit number and you want to find positive integers $a$ and $b$ such that $$a^b=ntag1$$ or prove that such numbers don't exists.



                  We have $$n<2^k$$ because $n$ is a $k$-bit number and so $$blt k$$



                  We can simply check for all possible $b$ if there is an $a$ such that $(1)$ holds. For given $b$ we can try to find $a$ by bisection. This bisection checks $O(log n)=O(k)$ different $a$. A check is the calculation of $a^b$. This can be achieved by multiplying powers of $a$ by $a$. These powers of $a$ are smaller than $n$. So we multiply $k$-bit numbers at most $b(lt k)$ times. A multiplication of two $k$-bit numbers needs $O(k^2)$ time. So all in all the algorithm needs $O(k^2)$ multiplications o $k$-bit numbers, which means $O(k^4)$ time.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 47 mins ago









                  miracle173miracle173

                  7,38022247




                  7,38022247



























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