Why does Solve lock up when trying to solve the equation

Solve[(x^2+y^2)+(x+y)==2^511 && x>0 && y>0,{x,y},Integers]

It works up 2^185, but at higher powers of 2, it seems to stop processing. The program says it's running, but there is no solution after running overnight. Running Mathematica 11.3 on Windows 32-bit OS.

  • 1
    $\begingroup$ You can format inline code and code blocks by selecting the code and clicking the {} button above the edit window. The edit window help button ? is useful for learning how to format your questions and answers. You may also find this meta Q&A helpful $\endgroup$ – Michael E2 Mar 10 at 18:27
  • 1
    $\begingroup$ I quickly get solutions for 2^257 (V11.3.0, macos). $\endgroup$ – Michael E2 Mar 10 at 18:29
  • $\begingroup$ Sorry, there was a typo - it should be 2^511 $\endgroup$ – user63373 Mar 10 at 19:21
  • 1
    $\begingroup$ Probably it's combinatorial blowup. n = 185 gives 32 solutions but n = 257 already gives 1024. Might be more work and memory to store the symbolics than your CPU can handle. $\endgroup$ – b3m2a1 Mar 10 at 19:38
  • 2
    $\begingroup$ @b3m2a1 I think the reasons probably have to do with number theory. n = 323 produces 8192 solutions in 2.3s and n = 325 produces only 128 solutions in 500s. The memory growth is quite low. I think for n = 511, you just have to wait long enough, and I can't predict how long that is. $\endgroup$ – Michael E2 Mar 10 at 19:57

Here's a guess: The Diophantine problem $$ x^2+y^2+x+y=a$$ is equivalent, via $u=2x+1,v=2y+1$ to finding the odd solutions to $$u^2+v^2=2+4a \,.$$ Whether Solve makes this transformation or not, solving the Pythagorean equation can be done from the prime factorization of $2+4a$. How long Solve takes thus might depend on how long it takes to factor $2+4a$.

This is not hard to verify:

Block[{FactorInteger = (Print["FactorInteger"[##]]; Abort[]) &},
 Print[2 + 4 2^325];
 Solve[(x^2 + y^2) + (x + y) == 2^325 && x > 0 && y > 0, {x, y}, Integers] // AbsoluteTiming



Well, it turns out it takes about 500 sec. to factor 2 + 4 * 2^325, which is about how long it takes the Solve above to run.

  • $\begingroup$ Thank you very much. This seems to be exactly what is happening. Much appreciated. $\endgroup$ – user63373 Mar 10 at 21:19
  • $\begingroup$ @user63373 You're welcome. :) $\endgroup$ – Michael E2 Mar 10 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.