Problem 750

First, $(a, y)=(1,1)$ is a solution. We assume therefore that $a, y \geq 2$. By looking modulo 9 and using the fact that the order of 7 modulo 9 is 3, we get that $y \equiv 0 \bmod 3$. We are thus looking for $p$ such that the order of 7 modulo $p$ is 3. In this case, we must have $p \mid 7^{3}-1$ and we see that $p=19$ works. Thus, $3^{2 a-1}+3^{a}+1 \equiv 1 \bmod 19$. (Alternatively, we could have directly said that since 3 divides $y$, 19 divides $7^{3}-1$ which divides $7^{y}-1$).

Therefore, $19 \mid 3^{a-1}+1$, so $3^{a-1} \equiv 18 \bmod 19$. The order of 3 modulo 19 is 18. We deduce that $a \equiv 10 \bmod 18$.

We then look for $p$ such that the order of 3 modulo $p$ is 18 or, to simplify, divides 18 if possible. We find that $p=7$ works (the order of 3 modulo 7 is 6). Modulo 7, we have $3^{19}+3^{10}+1 \equiv 0 \bmod 7$, which is absurd.