1. Find all integer solutions to the equation:
$$
x^{2}+x y+2 y^{2}=29
$$
Solving equation (1) for $x$, we get
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$

You get
$$
x=-3 \pm \sqrt{\frac{116-7 y^{2}}{2}} .
$$

Since $x$ is a real number, it must be that
$$
\begin{array}{l}
116-7 y^{2} \geqslant 0, \\
1 \quad \left\lvert\, \leqslant \sqrt{\frac{116}{7}}=4.07 \cdots .\right.
\end{array}
$$

Since $y$ is an integer, $y$ can only be $0, \pm 1$, $\pm 2, \pm 3, \pm 4$. However, only $y= \pm 4$ makes the expression under the square root a perfect square.

From equation (3), when $y=4$, $x=-1,-3$; and when $y=-4$, $x=3,1$.

Therefore, the integer solutions $(x, y)$ are $(1,-4)$, $(3,-4)$, $(-1,4)$, and $(-3,4)$.