Find all $x,y\in\mathbb{Z}^+$ such that $2014^x+11^x=y^2$

Question

Find all $x,y\in\mathbb{Z}^+$ such that $$2014^x+11^x=y^2$$ In my book it says that only solution is $(x,y)=(1,45)$, but solution is very complicated. They proved that $(x,y)=(1,45)$ is only solution using remainders of division, but I think that there must be an easier way. How would you solve this? What is the easiest way to solve equations like this?

Answer 1

This solution is 'using remainders of division', but I don't think it's very complicated.

Working $\mod 3$:

Recall that any square number is either $0$ or $1$ modulo $3$. This can easily be checked by squaring each of the three possible residues $\mod 3$ and seeing that you only ever get $0$ or $1$. Now:

\begin{align} 2014&\equiv 1&\mod 3\\ 11&\equiv-1&\mod 3 \end{align}

So our equation $$ 2014^x+11^x=y^2 $$ gives us that $$ 1+(-1)^x\in\{0,1\}\mod 3 $$ If $x$ is even, then $1+(-1)^x\equiv 2$, so $x$ must be odd.

Working $\mod 4$:

Recall that any square number is either $0$ or $1$ modulo $4$. Once again, this is easy to check by squaring the four possible residues $\mod 4$ and checking that you always get either $0$ or $1$. We have:

\begin{align} 2014&\equiv 2&\mod 4\\ 11&\equiv-1&\mod 4 \end{align}

So our equation $$ 2014^x+11^x=y^2 $$ gives us that $$ 2^x+(-1)^x\in\{0,1\}\mod 4 $$

Since $x$ is odd, $(-1)^x\equiv -1$, so we have: $$ 2^x-1\in\{0,1\}\mod 4 $$ If $x>1$, then $2^x\equiv0$, so $2^x-1\equiv -1$. Therefore, we must have $x=1$. Computation tells us that $2014^1+11^1=45^2$, so we have a unique solution $(1,45)$.