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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?

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    "Using remainders of division" -- in other words modular arithmetic -- is a straightforward and ubiquitous technique in elementary number theory. You probably won't find many arguments that are "easier" than that. – hmakholm left over Monica Dec 06 '14 at 17:02
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    To echo Henning's comment: Diophantine equations are frequently extremely hard to solve, and require techniques much harder than modular arithmetic. I doubt very much that there is an easier way to do this. – John Gowers Dec 06 '14 at 17:05
  • In fact, even polynomial equations can be unsolveable. Here it is worse because $x$ appears in the exponent. There is no general method to solve such equations. – Peter Dec 06 '14 at 17:09

1 Answers1

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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 $(1,45)$.

John Gowers
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