Show that $a^3+b^5=7^{7^{7^7}}$ has no solutions with $a,b\in \mathbb Z.$

Question

Show that $a^3+b^5=7^{7^{7^7}}$ has no solutions with $a,b\in \mathbb Z.$

I can't really get anywhere on this. I can't see how to find $7^{7^{7^{7}}}$ in any modulus other than $7$ or $6$, both of which are no use.

Any help is appreciated

Thank you.

Answer 1

Using Euler's theorem, you can compute $7^{7^{7^7}} \equiv 19 \mod 31$. However, $19$ is not expressible as the sum of a cube and a fifth power modulo $31$.

In other words, $a^3 + b^5 = 7^{7^{7^7}}$ has no solutions over $\mathbb{Z}/31\mathbb{Z}$, so it also has no solutions over $\mathbb{Z}$.

In general, with this type of question, it is useful to work modulo a prime $p$ such that the exponents involved divide $p-1$. The reason for this is that when $n \mid p-1$, $a^n$ cannot take many different values modulo $p$ (this is a consequence of Euler's theorem). Here $3$ and $5$ both divide $p-1=30$.

Edit: Euler's theorem can help you to compute $7^{7^{7^7}} \pmod {31}$ as follows. Since $(7,31)=1$ and $\phi(31)=30$, we have $7^{30} \equiv 1 \mod 31$. Therefore, $7^{7^{7^7}} \equiv 7^k \mod 31$, where $k$ is the remainder of $7^{7^7}$ upon division by $30$. In order to compute this, it suffices to compute $7^{7^7}$ modulo $2$, $3$ and $5$. For $7^{7^7} \pmod 5$ you can again use Euler's theorem to reduce this to $7^{\ell} \mod 5$, where $\ell$ is the remainder of $7^7$ upon division by $\phi(5)=4$.