Ramanujan-Nagell-ish diophantine equation

Question

The task is to find all $a, b \in \mathbb{Z}_+$ s.t.: $$2^a+17=b^4$$

I tried reducing modulo 17, but it doesn't really give much. Also a 4th power can $\equiv17$ for any $a$ big enough. Computer gives only the solution a=6, b=3.

Answer 1

Note that the order of $2$ $\pmod {17}$ is $8$, and that the non-zero fourth powers $\pmod {17}$ are $\{1,4,13,16\}$. The case where $b\equiv 0\pmod {17}$ yields no solutions.

Still working $\pmod {17}$ we see that the powers of $2$ which are fourth powers are $2^0, 2^2, 2^4,2^6$ from which we conclude that $a$ must be even. Write $a=2A$.

We then have $$17=b^4-2^{2A}=(b^2-2^A)(b^2+2^A)$$ So we must have $b^2-2^A=1, b^2+2^A=17$. This implies that $2b^2=18$ and we are done.