Intuition behind counterexample of Euler's sum of powers conjectureIntuition behind the Tamagawa numbers Is this a counterexample to a conjecture about independent domination in cartesian graph products?Goldbach's conjecture and Euler's idoneal numbersPossible counterexample to a theorem assuming Lang's conjectureCounterexample to Pólya's conjectureIntuition behind Kronecker's congruence?What is wrong with this counterexample to the Weak Bunyakovsky's conjecture and reformulation of Bunyakovsky's conjecture?Intuition behind the Riemann $zeta$ functional equationIntuition behind centralizers of Langlands parametersAlternating sum of powers

Intuition behind counterexample of Euler's sum of powers conjecture


Intuition behind the Tamagawa numbers Is this a counterexample to a conjecture about independent domination in cartesian graph products?Goldbach's conjecture and Euler's idoneal numbersPossible counterexample to a theorem assuming Lang's conjectureCounterexample to Pólya's conjectureIntuition behind Kronecker's congruence?What is wrong with this counterexample to the Weak Bunyakovsky's conjecture and reformulation of Bunyakovsky's conjecture?Intuition behind the Riemann $zeta$ functional equationIntuition behind centralizers of Langlands parametersAlternating sum of powers













20












$begingroup$


I was stunned when I first saw the article Counterexample to Euler's conjecture on sums of like powers by L. J. Lander and T. R. Parkin:.



enter image description here



How was it possible in 1966 to go through the sheer astronomical space of possibilities, on a CDC 6600 computer?




1) Did Lander and Parkin reveal their strategy?



2) How would you, using all the knowledge that was accessable until 1965, go to search for counter-examples, if you have access to a computer with 3 MegaFLOPS?




(As a comparison, todays home computers can have beyond 100 GigaFLOPS, using GPUs even TeraFLOPs)










share|cite|improve this question









$endgroup$







  • 4




    $begingroup$
    I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
    $endgroup$
    – Gerhard Paseman
    Mar 11 at 16:47










  • $begingroup$
    open access projecteuclid.org/euclid.bams/1183528522
    $endgroup$
    – Will Jagy
    Mar 11 at 17:09






  • 1




    $begingroup$
    There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
    $endgroup$
    – Michael Lugo
    Mar 11 at 18:56










  • $begingroup$
    With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
    $endgroup$
    – Wlod AA
    Mar 13 at 0:08
















20












$begingroup$


I was stunned when I first saw the article Counterexample to Euler's conjecture on sums of like powers by L. J. Lander and T. R. Parkin:.



enter image description here



How was it possible in 1966 to go through the sheer astronomical space of possibilities, on a CDC 6600 computer?




1) Did Lander and Parkin reveal their strategy?



2) How would you, using all the knowledge that was accessable until 1965, go to search for counter-examples, if you have access to a computer with 3 MegaFLOPS?




(As a comparison, todays home computers can have beyond 100 GigaFLOPS, using GPUs even TeraFLOPs)










share|cite|improve this question









$endgroup$







  • 4




    $begingroup$
    I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
    $endgroup$
    – Gerhard Paseman
    Mar 11 at 16:47










  • $begingroup$
    open access projecteuclid.org/euclid.bams/1183528522
    $endgroup$
    – Will Jagy
    Mar 11 at 17:09






  • 1




    $begingroup$
    There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
    $endgroup$
    – Michael Lugo
    Mar 11 at 18:56










  • $begingroup$
    With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
    $endgroup$
    – Wlod AA
    Mar 13 at 0:08














20












20








20


4



$begingroup$


I was stunned when I first saw the article Counterexample to Euler's conjecture on sums of like powers by L. J. Lander and T. R. Parkin:.



enter image description here



How was it possible in 1966 to go through the sheer astronomical space of possibilities, on a CDC 6600 computer?




1) Did Lander and Parkin reveal their strategy?



2) How would you, using all the knowledge that was accessable until 1965, go to search for counter-examples, if you have access to a computer with 3 MegaFLOPS?




(As a comparison, todays home computers can have beyond 100 GigaFLOPS, using GPUs even TeraFLOPs)










share|cite|improve this question









$endgroup$




I was stunned when I first saw the article Counterexample to Euler's conjecture on sums of like powers by L. J. Lander and T. R. Parkin:.



enter image description here



How was it possible in 1966 to go through the sheer astronomical space of possibilities, on a CDC 6600 computer?




1) Did Lander and Parkin reveal their strategy?



2) How would you, using all the knowledge that was accessable until 1965, go to search for counter-examples, if you have access to a computer with 3 MegaFLOPS?




(As a comparison, todays home computers can have beyond 100 GigaFLOPS, using GPUs even TeraFLOPs)







nt.number-theory ho.history-overview counterexamples






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 11 at 16:02









NicoDeanNicoDean

311324




311324







  • 4




    $begingroup$
    I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
    $endgroup$
    – Gerhard Paseman
    Mar 11 at 16:47










  • $begingroup$
    open access projecteuclid.org/euclid.bams/1183528522
    $endgroup$
    – Will Jagy
    Mar 11 at 17:09






  • 1




    $begingroup$
    There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
    $endgroup$
    – Michael Lugo
    Mar 11 at 18:56










  • $begingroup$
    With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
    $endgroup$
    – Wlod AA
    Mar 13 at 0:08













  • 4




    $begingroup$
    I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
    $endgroup$
    – Gerhard Paseman
    Mar 11 at 16:47










  • $begingroup$
    open access projecteuclid.org/euclid.bams/1183528522
    $endgroup$
    – Will Jagy
    Mar 11 at 17:09






  • 1




    $begingroup$
    There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
    $endgroup$
    – Michael Lugo
    Mar 11 at 18:56










  • $begingroup$
    With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
    $endgroup$
    – Wlod AA
    Mar 13 at 0:08








4




4




$begingroup$
I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
$endgroup$
– Gerhard Paseman
Mar 11 at 16:47




$begingroup$
I would compute a table of fifth powers (say 200 entries), then a table of sums of two fifth powers (say 20000 entries), and then do a search (say 2*10^8 tries). If I felt like using number theory I could eliminate some cases with considerations mod 5. Gerhard "Not Quite Pencil And Paper" Paseman, 2019.03.11.
$endgroup$
– Gerhard Paseman
Mar 11 at 16:47












$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
Mar 11 at 17:09




$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
Mar 11 at 17:09




1




1




$begingroup$
There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
$endgroup$
– Michael Lugo
Mar 11 at 18:56




$begingroup$
There appear to be some tidbits on how these sorts of searches are done in a 1967 paper: L. J. Lander, T. R. Parkin and J. L. Selfridge, A Survey of Equal Sums of Like Powers, Mathematics of Computation 21 (1967) 447-459 - ams.org/journals/mcom/1967-21-099/S0025-5718-1967-0222008-0/… The authors don't seem to specifically address the fifth-power case.
$endgroup$
– Michael Lugo
Mar 11 at 18:56












$begingroup$
With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
$endgroup$
– Wlod AA
Mar 13 at 0:08





$begingroup$
With today's computers, one could try $ale ble cle dle 1024. $ Did anybody?
$endgroup$
– Wlod AA
Mar 13 at 0:08











2 Answers
2






active

oldest

votes


















20












$begingroup$

The search strategy is laid out in a paper devoted to finding all "small" non-negative solutions of $$x_1^5+cdots x_n^5=y^5text with n leq 6$$




L. Lander & T. Parkin, "A counterexample to Euler's sum of powers
conjecture," Math. Comp., v. 21, 1967, pp. 101-103.




The one result was striking enough grab the title and get separate mention in the Bulletin.



The search was carried out for




  • $n=6$ and $y leq 100$ turning up $10$ primitive solutions of which two had $n=5$


  • $n=5$ and $y leq 250$ turning up $3$ more including the unexpected one for $n=4$


  • $n=4$ and $y leq 750$ turning up nothing else new.

That last search covers more than $5000$ times as many cases as going to $133^5.$



The length of the paper hints that the method is fairly simple. Space must have been limited because a table of fifth powers was maintained but not a table of sums $a^5+b^5$ for a meet in the middle attack.






share|cite|improve this answer











$endgroup$




















    16












    $begingroup$

    Even simply generating all quadruples $(a, b, c, d)$ with $1 le a le b le c le d le 133$ should work fine. There are only about 13 million such quadruples. For each, we need to add together the fifth powers (which can be looked up in a small table) and check whether the result is another fifth power, which we can do by binary search in a table of fifth powers (of size, say, 200 or so). I'm not familiar with the CDC 6600 but it seems like a direct implementation of this kind would finish in under an hour.



    There are faster algorithms possible, using the "meet-in-the-middle" method. For example, record all the sums $a^5 + b^5$ for $1 le a le b le 200$ or so, and sort them. Now, for each $1 le c le d le e$, compute $e^5 - c^5 - d^5$ and check whether it is in this table. This algorithm is cubic (up to log factors) in the size of the eventual counterexample, rather than quartic. However, the actual counterexample in this case is small enough that this kind of method seems to be unnecessary.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
      $endgroup$
      – Michael Lugo
      Mar 11 at 18:50






    • 6




      $begingroup$
      One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
      $endgroup$
      – Greg Martin
      Mar 11 at 22:17







    • 1




      $begingroup$
      Using $x^5=x /bmod 30$ gives a constant factor of $30.$
      $endgroup$
      – Aaron Meyerowitz
      Mar 12 at 7:27










    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    20












    $begingroup$

    The search strategy is laid out in a paper devoted to finding all "small" non-negative solutions of $$x_1^5+cdots x_n^5=y^5text with n leq 6$$




    L. Lander & T. Parkin, "A counterexample to Euler's sum of powers
    conjecture," Math. Comp., v. 21, 1967, pp. 101-103.




    The one result was striking enough grab the title and get separate mention in the Bulletin.



    The search was carried out for




    • $n=6$ and $y leq 100$ turning up $10$ primitive solutions of which two had $n=5$


    • $n=5$ and $y leq 250$ turning up $3$ more including the unexpected one for $n=4$


    • $n=4$ and $y leq 750$ turning up nothing else new.

    That last search covers more than $5000$ times as many cases as going to $133^5.$



    The length of the paper hints that the method is fairly simple. Space must have been limited because a table of fifth powers was maintained but not a table of sums $a^5+b^5$ for a meet in the middle attack.






    share|cite|improve this answer











    $endgroup$

















      20












      $begingroup$

      The search strategy is laid out in a paper devoted to finding all "small" non-negative solutions of $$x_1^5+cdots x_n^5=y^5text with n leq 6$$




      L. Lander & T. Parkin, "A counterexample to Euler's sum of powers
      conjecture," Math. Comp., v. 21, 1967, pp. 101-103.




      The one result was striking enough grab the title and get separate mention in the Bulletin.



      The search was carried out for




      • $n=6$ and $y leq 100$ turning up $10$ primitive solutions of which two had $n=5$


      • $n=5$ and $y leq 250$ turning up $3$ more including the unexpected one for $n=4$


      • $n=4$ and $y leq 750$ turning up nothing else new.

      That last search covers more than $5000$ times as many cases as going to $133^5.$



      The length of the paper hints that the method is fairly simple. Space must have been limited because a table of fifth powers was maintained but not a table of sums $a^5+b^5$ for a meet in the middle attack.






      share|cite|improve this answer











      $endgroup$















        20












        20








        20





        $begingroup$

        The search strategy is laid out in a paper devoted to finding all "small" non-negative solutions of $$x_1^5+cdots x_n^5=y^5text with n leq 6$$




        L. Lander & T. Parkin, "A counterexample to Euler's sum of powers
        conjecture," Math. Comp., v. 21, 1967, pp. 101-103.




        The one result was striking enough grab the title and get separate mention in the Bulletin.



        The search was carried out for




        • $n=6$ and $y leq 100$ turning up $10$ primitive solutions of which two had $n=5$


        • $n=5$ and $y leq 250$ turning up $3$ more including the unexpected one for $n=4$


        • $n=4$ and $y leq 750$ turning up nothing else new.

        That last search covers more than $5000$ times as many cases as going to $133^5.$



        The length of the paper hints that the method is fairly simple. Space must have been limited because a table of fifth powers was maintained but not a table of sums $a^5+b^5$ for a meet in the middle attack.






        share|cite|improve this answer











        $endgroup$



        The search strategy is laid out in a paper devoted to finding all "small" non-negative solutions of $$x_1^5+cdots x_n^5=y^5text with n leq 6$$




        L. Lander & T. Parkin, "A counterexample to Euler's sum of powers
        conjecture," Math. Comp., v. 21, 1967, pp. 101-103.




        The one result was striking enough grab the title and get separate mention in the Bulletin.



        The search was carried out for




        • $n=6$ and $y leq 100$ turning up $10$ primitive solutions of which two had $n=5$


        • $n=5$ and $y leq 250$ turning up $3$ more including the unexpected one for $n=4$


        • $n=4$ and $y leq 750$ turning up nothing else new.

        That last search covers more than $5000$ times as many cases as going to $133^5.$



        The length of the paper hints that the method is fairly simple. Space must have been limited because a table of fifth powers was maintained but not a table of sums $a^5+b^5$ for a meet in the middle attack.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 12 at 7:33

























        answered Mar 11 at 18:58









        Aaron MeyerowitzAaron Meyerowitz

        24.2k13288




        24.2k13288





















            16












            $begingroup$

            Even simply generating all quadruples $(a, b, c, d)$ with $1 le a le b le c le d le 133$ should work fine. There are only about 13 million such quadruples. For each, we need to add together the fifth powers (which can be looked up in a small table) and check whether the result is another fifth power, which we can do by binary search in a table of fifth powers (of size, say, 200 or so). I'm not familiar with the CDC 6600 but it seems like a direct implementation of this kind would finish in under an hour.



            There are faster algorithms possible, using the "meet-in-the-middle" method. For example, record all the sums $a^5 + b^5$ for $1 le a le b le 200$ or so, and sort them. Now, for each $1 le c le d le e$, compute $e^5 - c^5 - d^5$ and check whether it is in this table. This algorithm is cubic (up to log factors) in the size of the eventual counterexample, rather than quartic. However, the actual counterexample in this case is small enough that this kind of method seems to be unnecessary.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
              $endgroup$
              – Michael Lugo
              Mar 11 at 18:50






            • 6




              $begingroup$
              One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
              $endgroup$
              – Greg Martin
              Mar 11 at 22:17







            • 1




              $begingroup$
              Using $x^5=x /bmod 30$ gives a constant factor of $30.$
              $endgroup$
              – Aaron Meyerowitz
              Mar 12 at 7:27















            16












            $begingroup$

            Even simply generating all quadruples $(a, b, c, d)$ with $1 le a le b le c le d le 133$ should work fine. There are only about 13 million such quadruples. For each, we need to add together the fifth powers (which can be looked up in a small table) and check whether the result is another fifth power, which we can do by binary search in a table of fifth powers (of size, say, 200 or so). I'm not familiar with the CDC 6600 but it seems like a direct implementation of this kind would finish in under an hour.



            There are faster algorithms possible, using the "meet-in-the-middle" method. For example, record all the sums $a^5 + b^5$ for $1 le a le b le 200$ or so, and sort them. Now, for each $1 le c le d le e$, compute $e^5 - c^5 - d^5$ and check whether it is in this table. This algorithm is cubic (up to log factors) in the size of the eventual counterexample, rather than quartic. However, the actual counterexample in this case is small enough that this kind of method seems to be unnecessary.






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
              $endgroup$
              – Michael Lugo
              Mar 11 at 18:50






            • 6




              $begingroup$
              One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
              $endgroup$
              – Greg Martin
              Mar 11 at 22:17







            • 1




              $begingroup$
              Using $x^5=x /bmod 30$ gives a constant factor of $30.$
              $endgroup$
              – Aaron Meyerowitz
              Mar 12 at 7:27













            16












            16








            16





            $begingroup$

            Even simply generating all quadruples $(a, b, c, d)$ with $1 le a le b le c le d le 133$ should work fine. There are only about 13 million such quadruples. For each, we need to add together the fifth powers (which can be looked up in a small table) and check whether the result is another fifth power, which we can do by binary search in a table of fifth powers (of size, say, 200 or so). I'm not familiar with the CDC 6600 but it seems like a direct implementation of this kind would finish in under an hour.



            There are faster algorithms possible, using the "meet-in-the-middle" method. For example, record all the sums $a^5 + b^5$ for $1 le a le b le 200$ or so, and sort them. Now, for each $1 le c le d le e$, compute $e^5 - c^5 - d^5$ and check whether it is in this table. This algorithm is cubic (up to log factors) in the size of the eventual counterexample, rather than quartic. However, the actual counterexample in this case is small enough that this kind of method seems to be unnecessary.






            share|cite|improve this answer









            $endgroup$



            Even simply generating all quadruples $(a, b, c, d)$ with $1 le a le b le c le d le 133$ should work fine. There are only about 13 million such quadruples. For each, we need to add together the fifth powers (which can be looked up in a small table) and check whether the result is another fifth power, which we can do by binary search in a table of fifth powers (of size, say, 200 or so). I'm not familiar with the CDC 6600 but it seems like a direct implementation of this kind would finish in under an hour.



            There are faster algorithms possible, using the "meet-in-the-middle" method. For example, record all the sums $a^5 + b^5$ for $1 le a le b le 200$ or so, and sort them. Now, for each $1 le c le d le e$, compute $e^5 - c^5 - d^5$ and check whether it is in this table. This algorithm is cubic (up to log factors) in the size of the eventual counterexample, rather than quartic. However, the actual counterexample in this case is small enough that this kind of method seems to be unnecessary.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 11 at 16:42









            Reid BartonReid Barton

            19.2k152110




            19.2k152110











            • $begingroup$
              The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
              $endgroup$
              – Michael Lugo
              Mar 11 at 18:50






            • 6




              $begingroup$
              One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
              $endgroup$
              – Greg Martin
              Mar 11 at 22:17







            • 1




              $begingroup$
              Using $x^5=x /bmod 30$ gives a constant factor of $30.$
              $endgroup$
              – Aaron Meyerowitz
              Mar 12 at 7:27
















            • $begingroup$
              The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
              $endgroup$
              – Michael Lugo
              Mar 11 at 18:50






            • 6




              $begingroup$
              One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
              $endgroup$
              – Greg Martin
              Mar 11 at 22:17







            • 1




              $begingroup$
              Using $x^5=x /bmod 30$ gives a constant factor of $30.$
              $endgroup$
              – Aaron Meyerowitz
              Mar 12 at 7:27















            $begingroup$
            The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
            $endgroup$
            – Michael Lugo
            Mar 11 at 18:50




            $begingroup$
            The fact that the authors say they used "direct search" and don't elaborate suggests, to me, that they didn't do anything fancy, and probably used this approach.
            $endgroup$
            – Michael Lugo
            Mar 11 at 18:50




            6




            6




            $begingroup$
            One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
            $endgroup$
            – Greg Martin
            Mar 11 at 22:17





            $begingroup$
            One could easily gain a constant factor of 2ish by noting, for example, that all 5th powers are in $-1,0,1$ modulo $11$; this rules out lots of possible quadruples.
            $endgroup$
            – Greg Martin
            Mar 11 at 22:17





            1




            1




            $begingroup$
            Using $x^5=x /bmod 30$ gives a constant factor of $30.$
            $endgroup$
            – Aaron Meyerowitz
            Mar 12 at 7:27




            $begingroup$
            Using $x^5=x /bmod 30$ gives a constant factor of $30.$
            $endgroup$
            – Aaron Meyerowitz
            Mar 12 at 7:27

















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