How can I solve this equation in $\mathbb Z$?

Question

$x^{2}+16x+55=3^{y^{2}-2y}$ Could you tell me how to solve this equation, please? I tried to draw the graphic of the functions but I didn't see how to solve it.

Answer 1

Factoring gives $(x+5)(x+11)=3^{y^2-2y}$, hence \begin{align} &x+5=3^u& &x+11=3^v \end{align} for some $u,v$ with $0\leq u<v$. Then \begin{align} 6 &=(x+11)-(x+5)\\ &=3^u(3^{v-u}-1) \end{align} from which $u=1$ and $v-u=1$, hence $v=2$. Consequently, $x=-2$ and $y^2-2y=3$ from which $y=3$ or $y=-1$.

Answer 2

Hint

$$3^{y^2-2y}=(x+11)(x+5)$$

As the left side is divisible by $3$ for $y^2-2y>1\iff(y-1)^2>2\implies y\ge3$ as $y$ is an integer

and as $x+11-(x+5)=6$

Either both or none is divisible by $3$

$$3^{y^2-2y-2}=\dfrac{x+11}3\cdot\dfrac{x+5}3$$

As exactly one of the multiplicands can be divisible by $3,$ the other must be $\pm1$

Check for $y\le2$ separately