Finding the solutions of an equation in positive intgers

Question

We are asked to find all positive integer solutions of the equation $$x^7+7=y^2$$ or to show that it does not have any solutions. It is clear that $x$ can not be even. So $x$ is odd and $y$ is even (In fact $x$ is of the form $4k+1$). Adding $121$ to both sides of the equation, we get $x^7+128=y^2+121$. Hence $$(x+2)(x^6-2x^5+4x^4-8x^3+16x^2-32x+64)=y^2+121.$$ Since the second expression on LHS is of the form $4k+3$, it does have a prime divisor $p$ with the same form. So $$p|y^2+121$$, and from a previous lemma, we can conclude that $p|11$ and $p|y$, from which we obtain $p=11$ and $484| x^7+7$. If we proceed further, for $x^7+7$ to be divisible by 11, $x$ has to be of the form $11k+9$, where $k$ itself is shown to be of the form $11t+10$ giving us $x$ is of the form $121t+119$. How to continue in this way to show that such an equation does not have any solutions? Any help will be appreciated. Thank you.

Answer 1

I’m turning my comment into an answer.

Let $x$ be an integer such that $11|x^7+128$. Then clearly $11$ and $x$ are coprime and $11|(x^7)^3+128^3=x^{21}+2^{21}$. By (Pierre) Fermat’s little theorem, $11|x^{21}-x$, $11|2^{21}-2$ and thus $11|x+2$. Then write $x=11y-2$, then $\frac{x^7+128}{x+2}=\frac{(11y-2)^7+2^7}{11y}=\frac{11^2Ny+11\times 14y}{11y}=11N+14$ isn’t divisible by $11$.

In other words, for any integer $x$, $\frac{x^7+128}{x+2}$ is coprime to $11$. But we know from the OP that if $x^7+7=y^2$, then $11$ divides $\frac{x^7+128}{x+2}$, and we get a contradiction.