3. Solve in the set of integers the equation $x^{2}+16 x+55=3^{y^{2}-2 y}$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. 

Note: The note itself is not part of the translation but is provided to clarify the instruction. Here is the translation as requested:

3. Solve in the set of integers the equation $x^{2}+16 x+55=3^{y^{2}-2 y}$.
Subject 3. Solve in the set of integers the equation $x^{2}+16 x+55=3^{y^{2}-2 y}$.

Prof. Mihaela Berindeanu, Bucharest

| Details of solution | Associated grading |
| :--- | :--- |
| The equation can also be written as $(x+5)(x+11)=3^{y^{2}-2 y}$. Since $(x+5)(x+11) \in \mathbb{Z}$, it follows that |  |
| $3^{y^{2}-2 y} \in \mathbb{N}^{*}$, hence $y^{2}-2 y \in \mathbb{N}^{*}$. | $\mathbf{2 p}$ |
| This means that the numbers $x+5$ and $x+11$ are simultaneously positive or simultaneously |  |
| negative integers. We deduce that $\|x+11\|-\mid x+5 \\|=6$. |  |
| Since the numbers $\|x+5\|$ and $\|x+11\|$ are simultaneously positive integers, we have |  |
| $\|x+5\|=3^{a}, a \in \mathbb{N}$ and $\|x+11\|=3^{b}, b \in \mathbb{N}$, where $a+b=y^{2}-2 y$. If, for example, $b>a$, |  |
| then $3^{b}-3^{a}=6$, equivalent to $3^{a}\left(3^{b-a}-1\right)=6$. We deduce that $3^{a} \mid 6$, so $a \in\{0,1\}$. | 2p |
| Only $a=1$ works, from which $b=2$. |  |
| Therefore, $y^{2}-2 y=3$ with solutions $y=3, y=-1$. |  |
| Thus, $(x+5)(x+11)=27$. From $x+5=3, x+11=9$ we get $x=-2$. If $x+5=-9, x+11=-3$ we get $x=-14$. In the end, $(x, y) \in\{(-2,3),(-2,-1),(-14,3),(-14,-1)\}$. |  |