Intuition behind counterexample of Euler's sum of powers conjecture
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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:.
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
asked 7 hours ago
NicoDeanNicoDean
211322
$endgroup$
add a comment |
$begingroup$
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:.
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
asked 7 hours ago
NicoDeanNicoDean
211322
$endgroup$
2
$begingroup$
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.
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– Gerhard Paseman
6 hours ago
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open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
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
4 hours ago
add a comment |
$begingroup$
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:.
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
asked 7 hours ago
NicoDeanNicoDean
211322
$endgroup$
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:.
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
nt.number-theory ho.history-overview counterexamples
asked 7 hours ago
NicoDeanNicoDean
211322
asked 7 hours ago
NicoDeanNicoDean
211322
asked 7 hours ago
NicoDeanNicoDean
211322
asked 7 hours ago
NicoDeanNicoDean
211322
asked 7 hours ago
NicoDeanNicoDean
211322
211322
2
$begingroup$
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
6 hours ago
$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
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
4 hours ago
add a comment |
2
$begingroup$
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
6 hours ago
$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
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
4 hours ago
2
2
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
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
4 hours ago
$begingroup$
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
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $133^5.$
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
$endgroup$
add a comment |
$begingroup$
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.
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
$endgroup$
$begingroup$
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
4 hours ago
1
$begingroup$
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
1 hour ago
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
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 $133^5.$
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
$endgroup$
add a comment |
$begingroup$
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 $133^5.$
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
$endgroup$
add a comment |
$begingroup$
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 $133^5.$
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
$endgroup$
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 $133^5.$
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
edited 4 hours ago
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
answered 4 hours ago
Aaron MeyerowitzAaron Meyerowitz
24.1k13288
24.1k13288
add a comment |
add a comment |
$begingroup$
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.
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
$endgroup$
$begingroup$
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
4 hours ago
1
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
$endgroup$
$begingroup$
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
4 hours ago
1
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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.
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
$endgroup$
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.
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
answered 7 hours ago
Reid BartonReid Barton
18.7k150106
18.7k150106
$begingroup$
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
4 hours ago
1
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
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
4 hours ago
1
$begingroup$
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
1 hour ago
$begingroup$
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
4 hours ago
$begingroup$
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
4 hours ago
1
1
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
add a comment |
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$begingroup$
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
6 hours ago
$begingroup$
open access projecteuclid.org/euclid.bams/1183528522
$endgroup$
– Will Jagy
6 hours ago
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
4 hours ago