Show that $x^3 + y^3 + z^3 + t^3 = 1999$ has infinitely many integer solutions.

I have not been able to find a single solution to this equation. With some trial I think there does not exist a solution with all of them positive. Can you please help me proceed?


  • $\begingroup$ $(2n+14)^3-(2n+23)^3-(3n+26)^3+(3n+30)^3=18n+1$ and $n=111$ gives a solution. $\endgroup$
    – almagest
    May 9, 2016 at 14:48
  • $\begingroup$ If the numbers are restricted to positive integers, then, of course, there are only a finite number of solutions. I think that they meant to domain to be all integers. $\endgroup$ May 9, 2016 at 14:49
  • $\begingroup$ @StevenGregory Yes. $\endgroup$ May 9, 2016 at 14:50
  • $\begingroup$ @almagest How did you derive this? $\endgroup$ May 9, 2016 at 14:51
  • 1
    $\begingroup$ @almagest, googling on "1999 has infinitely many integer solutions" gives various putnam and olympiad hits. $\endgroup$ May 9, 2016 at 15:01

1 Answer 1


Look for solutions $$10-b,10+b,-\frac{1}{2}(d+1),\frac{1}{2}(d-1)$$ The sum of these numbers cubed is $2000+60b^2-\frac{1}{4}(3d^2+1)$, so we need $240b^2-3d^2-1=-4$ or $d^2-80b^2=1$. It is easy to see that has the solution $d=9,b=1$. Now if $d,b$ is a solution then $(9d+80b)^2-80(9b+d)^2=d^2-80b^2$, so $d'=9d+80b,b'=9b+d$ is another solution. Note that $d$ will always be odd.

Thus we have infinitely many integer solutions to the original equation.

[This was a question in the Bulgarian National Olympiad for 1999.]


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.