11.5. Find all solutions in natural numbers of the equation: $2^{x}+3^{y}=z^{2}$.
Answer: $x=4, y=2, z=5$.

Solution. 1) The power of two is not divisible by 3, so the left side of the equation is not divisible by 3, hence $z^{2}$ and $z$ are not divisible by 3. Therefore, the remainder of the division of the right side of the equation by 3 is 1.

2) Consider the left side modulo 3, it is equal to $(-1)^{x}$ and by 1) it is comparable to 1. Therefore, $x=2x_{1}, x_{1} \geq 1$ - even.
3) Then $3^{y}=z^{2}-2^{2x_{1}}=(z-2^{x_{1}})(z+2^{x_{1}})$, from which $z+2^{x_{1}}=3^{a}$ and $z-2^{x_{1}}=3^{b}$ for some non-negative integers $a, b$ such that $a+b=y$. Subtracting the second from the first, we get $2^{x_{1}+1}=3^{a}-3^{b}=3^{b}(3^{a-b}-1)$, from which $b=0$, because the power of two is not divisible by 3.
4) Factorize $2^{x_{1}+1}=3^{a}-1=(3-1)(3^{a-1}+\ldots+3^{1}+1)$, from which $2^{x_{1}}=3^{a-1}+\ldots+3^{1}+1$. On the right, there are $a$ odd terms, on the left - the even number $2^{x_{1}}, x_{1} \geq 1$ ? therefore, $a=2a_{1}$ - even.
5) Now $2^{x_{1}+1}=3^{2a_{1}}-1=(3^{a_{1}}-1)(3^{a_{1}}+1)$ and the numbers $3^{a_{1}}-1$ and $3^{a_{1}}+1$ are (analogously to point 3)) powers of two, differing by 2. Subtracting them from each other, we get that the smaller one is 2, and the larger one is 4. Therefore, $2^{x_{1}+1}=2 \cdot 4=8$, from which $x_{1}=2, x=2x_{1}=4$. From point 4 we have $3^{a}-1=2^{x_{1}+1}=8$, so $y=a=2$ and $z=5$.

Criterion. Only the answer: 1 point.

Evenness of $x=2x_{1}, x_{1} \geq 1: 2$ points.

Evenness of $a=2a_{1}: 2$ points.