# Finding integer solutions to $3^y = x^2 + 56$

The original question is asking for the following:

Find all ordered pairs of positive integers $(x, y)$ for which $3^y = x^2 + 56$.

I've found a solution I believe to be the only solution, but I'm struggling to prove it's the only solution. $(5,4)$

When I see a problem like this, I initially think to use modular arithmetic to find the pattern of the answer. For example, taking each side of the equation $\mod 4$ shows that $x$ must be odd and $y$ must be even. $\mod 3$ shows that $x$ is not divisible by $3$.

Using a spreadsheet and running all $\mod \$ from $2$ to $81$ and looking at the remainders for $x = 1$ to $250$ shows that $x = 5$ is the only possible solution, but obviously this is not a proof. I'm struggling to find the steps necessary to prove this answer.

I've also simplified the equation to have $y=\frac{\ln(x^2+56)}{\ln(3)}$, but I don't know how to show the only time $y$ is an integer is when $x=5$. (Perhaps I am wrong and more solutions exist). Any help would be appreciated.

Since $y>0$, $x$ must be odd. Also, as you said, since $3^y\equiv_4(-1)^y\equiv_4 x^2+56\equiv_4 0$, $y$ must be even. Notice that $$3^{y}-x^2=(3^{\frac{y}{2}}-x)(3^{\frac{y}{2}}+x)=56=2^3\cdot7$$
So there are only finitely many solutions, since $3^{\frac{y}{2}}+x>0$ divides $56$ (and take up atmost 8 values, since 56 has 8 divisors).
Generally, if $a\cdot b=c$ and $a\geq b$ the smallest value for $a$ is $\sqrt{c}$ (one can prove this by contradiction). Since $3^{\frac{y}{2}}+x$ is the larger factor, it is atleast $\sqrt{56}\approx 7.48>7$. So you only have to check the cases: $$3^{\frac{y}{2}}+x=14=2\cdot 7$$ $$3^{\frac{y}{2}}+x=28=2^2\cdot 7$$ $$3^{\frac{y}{2}}+x=56=2^3\cdot7$$
The first case is your solution $(5,4)$. I'm not sure how one would write the rest in a formal proof, but I'll guess using a computer (and refering to the calculations) is OK, since there are only finitely many solutions.