Find a non-trivial solution for the Diophantine Equation $17a^4 + 5b^4 = 35c^4$, or show that no non-trivial solutions exist

Question

This is a problem on my practice exam for number theory, and we haven't had an example like this in class yet. The question is looking for a solution in $\mathbb{Z}$ for $a,b,c \in \mathbb{Z}$. I've tried reducing everything (mod 5), but I didn't really figure anything out from that, and I'm not seeing many options for quadratic reciprocity either...

Answer 1

Modulo $5$, we can easily conclude that $5\mid a$. Dividing by $5$, and looking mod $5$ again, we can conclude that $5\mid b$ and $5\mid c$.

The result, that only $(0,0,0)$ is a solution, follows by infinite descent.

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There also appears to be a descent argument modulo $4$, and another modulo $17$.

Answer 2

$\bmod 17\!:\:\:\: 17a^4+5b^4-35c^4\equiv 5b^4-c^4\equiv 0\,\Rightarrow\, 17\mid b,c\,\Rightarrow\, 17\mid a$,

since $5b^4\equiv\{0,3,5,12,14\}$ (WA), $\ c^4\equiv\{0,1,4,13,16\}\pmod{\!17}$ (WA).

So $(a/17,b/17,c/17)$ gives another integer solution for arbitrary $(a,b,c)$, which is only possible when $(a,b,c)=(0,0,0)$ (infinite descent). As Slade said, mod $4,5$ too give infinite descent proofs.