231*. $x^{2}+x y+y^{2}+x+y-5=0$, if one solution is known: $x=1 ; y=1$.
231. Let $x=x_{1}+1$ and $y=y_{1}+1$, we get the equation $x_{1}^{2}+y_{1}^{2}+x_{1} y_{1}+4 x_{1}+4 y_{1}=0, x_{1}=0, y_{1}=0$.

Let $\frac{x_{1}}{y_{1}}=\frac{m}{n}$, where $m$ and $n$ are coprime numbers; we get $\left(m^{2}+m n+n^{2}\right) x_{1}^{2}+4 m(m+n) x_{1}=0$; discarding the obvious solution $x_{1}=0$, we find

$$
x_{1}=-\frac{4 m(m+n)}{m^{2}+m n+n^{2}}, \quad y_{1}=-\frac{4 n(m+n)}{m^{2}+m n+n^{2}}
$$

We will show that the numbers $m(m+n)$ and $m^{2}+m n+n^{2}$ are coprime. Indeed, any common prime divisor of the numbers $m^{2}+m n$ and $m^{2}+m n+n^{2}$ must be a divisor of their difference, which is equal to $n^{2}$, and therefore of the number $n$; but then this divisor must be a part of the number $m$, which is impossible since $m$ and $n$ are coprime. Based on this, we conclude that $m^{2}+m n+n^{2}$ must be a divisor of the number 4. Since $m^{2}+m n+n^{2}=\left(m+\frac{n}{2}\right)^{2}+\frac{3}{4} n^{2}>0$, we consider the following cases:

1) $m^{2}+m n+n^{2}=1$, hence $m= \pm 1, n=0$; therefore, $x_{1}=-4, y_{1}=0 ; x=-3, y=1$.
2) $m^{2}+m n+n^{2}=2$, which is impossible, since

$$
m=\frac{-n \pm \sqrt{8-3 n^{2}}}{2} ; \text { hence }|n| \leqslant \sqrt{\frac{8}{3}}
$$

therefore, $|n|=1$, but when $|n|=1$ the number $8-3 n^{2}$ is not a perfect square.
3) $m^{2}+m n+n^{2}=4$, hence $|n| \leqslant \sqrt{\frac{16}{3}}, \quad|n|=+1$; $|n|=+2$.

When $n= \pm 2 m=0, x_{1}=0, y_{1}=-4 ;$ therefore, $x=1, y=-3$.

When $|n|=1 m$ is an irrational number.

Answer. $(1,1),(1,-3),(-3,1)$.