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

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?

Thanks.

• $(2n+14)^3-(2n+23)^3-(3n+26)^3+(3n+30)^3=18n+1$ and $n=111$ gives a solution. May 9, 2016 at 14:48
• 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. May 9, 2016 at 14:49
• @StevenGregory Yes. May 9, 2016 at 14:50
• @almagest How did you derive this? May 9, 2016 at 14:51
• @almagest, googling on "1999 has infinitely many integer solutions" gives various putnam and olympiad hits. May 9, 2016 at 15:01

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.