This is one of the problems that my teacher gave out, in which he mentioned that $(x, y, z, t) = (2, 2, 5, 1)$ is the only solution to this equation. However, I haven't been able to prove this at all or find any solutions for this problem, so far I have only found out that $x, y$ are both even and $z$ is divisible by $5$.
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2Please work out how you found this out , this may give other users an idea how to proceed. – Peter Dec 08 '23 at 16:16
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2Seems that positive integers are reuqired since otherwise $x=y=z=t=0$ would also be a solution. – Peter Dec 08 '23 at 16:17
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Using mod 3 $y$ is even, using mod 5 $z, t$ have the same parity, using mod 8 $x$ is even, using mod 25 $z$ is divisible by 5. – Dũng Nguyễn Dec 08 '23 at 16:27
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1Right, it works this way, compare with this post. – Dietrich Burde Dec 08 '23 at 16:27
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The equation is solved here, in Theorem $4.22$ on page $279$, by Brenner and Foster. The only nonnegative integral solutions are $$ (0,0,0,0), \;(1,0,2,1),\; (2,2,5, 1). $$ One of the intermediate results is Lemma $4.224$, where it is shown that $$ (x,y,z,t)\equiv (2,2,5,1) \bmod 60. $$
Dietrich Burde
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