161. Solve the equation in positive integers

$$
x^{2 y}+(x+1)^{2 y}=(x+2)^{2 y}
$$
161. For $y=1$ we arrive at the quadratic equation

$$
x^{2}+(x+1)^{2}=(x+2)^{2}, \text { or } x^{2}-2 x-3=0
$$

which has a single positive integer root $x=3$ (the second root of the equation $x=-1$ is negative). Let now $y>1$. Note that since $x^{2 y}$ and $(x+2)^{2 y}$ are numbers of the same parity, the number $(x+1)^{2 y} \left[=(x+2)^{2 y}-x^{2 y}\right]$ will be even; therefore, the number $x+1=2 x_{1}$ is even. Further, we have

$$
\begin{aligned}
\left(2 x_{1}\right)^{2 y}=(x+1)^{2 y}=(x+2)^{2 y}-x^{2 y} & = \\
& =\left(2 x_{1}+1\right)^{2 y}-\left(2 x_{1}-1\right)^{2 y}
\end{aligned}
$$

Expanding the expressions on the right using the binomial theorem, we get:

$$
\left(2 x_{1}\right)^{2 y}=2\left[C_{2 y}^{1}\left(2 x_{1}\right)^{2 y-1}+C_{2 y}^{3}\left(2 x_{1}\right)^{2 y-3}+\ldots+C_{2 y}^{1}\left(2 x_{1}\right)\right]
$$

which, isolating the last term of the right-hand side in the left-hand side, can be rewritten as:

$2 \cdot 2 y \cdot 2 x_{1}=$

$=\left(2 x_{1}\right)^{2 y}-2 \cdot 2 y\left(2 x_{1}\right)^{2 y-1}-2 \cdot C_{2 y}^{3}\left(2 x_{1}\right)^{2 y-3}-\ldots-2 \cdot C_{2 y}^{3}\left(2 x_{1}\right)^{s}$

(Note that since $y \geqslant 2$, then $2 y \geqslant 4$). But all terms on the right-hand side obviously divide by $\left(2 x_{1}\right)^{3}$, from which it follows that $y$ must divide by $x_{1}^{2}$.

Further, dividing both sides of equation (*) by $\left(2 x_{1}\right)^{2 y}$, we get:

$$
1=\left(1+\frac{1}{2 x_{1}}\right)^{2 y}-\left(1-\frac{1}{2 x_{1}}\right)^{2 y}
$$

from which it follows that $\left(1+\frac{1}{2 x_{1}}\right)^{2 y}=1+\left(1-\frac{1}{2 x_{1}}\right)^{2 y}1+\frac{y}{x_{1}}
$$

and, therefore,

$$
1+\frac{y}{x_{1}}1$ and all solutions of the equation are exhausted by the one found above: $y=1, x=3$.