<s>[INST] Proof check modulo arithmetic what is wrong with this proof,
so I am supposed to show $7x^2-15y^2=1$ has no integer solutions...
so since $7x^2=1+3(5y^2)$ so $7x^2\equiv 1 \pmod{3}$
hence in mod $3$,  $x\equiv 0, 1$ or $2$ so $x^2\equiv 0, 1$ and $7x^2\equiv0$ or $7$
and since $7$ is congruent to $1 \bmod3$ I end up with solutions.
I also tried it with mod $5$ since $15y^2=5(3y^2)$ and I ended up with no solution so why not in case above?
 [/INST] There are also no integer solutions to $$ 13 x^2 - 17 y^2 = 1.  $$
We know that there are rational solutions, therefore $p$-adic, as we can solve
$$  13 x^2 - 17 y^2 = 25  $$
in integers.
$$  13 \cdot 7^2 - 17 \cdot 6^2 = 637 - 612 = 25  $$
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