Let's determine the positive integer solutions for the equation

$$
29 x-23 y=123
$$

(For more practice on this topic, see the article entitled "Integer Solutions of Linear Equations" in this issue.)
Following the learned algorithm,

$$
y=\frac{29 x-123}{23}=x-5+\frac{6 x-8}{23}=x-5+u
$$

where $u$ is an integer and

$$
6 x-8=23 u
$$

thus

$$
x=\frac{23 u+8}{6}=3 u+1+\frac{5 u+2}{6}=3 u+1+v
$$

from which

$$
u=\frac{6 v-2}{5}=v+\frac{v-2}{5}=v+t
$$

and thus

$$
\begin{gathered}
v=5 t+2, \quad u=v+t=6 t+2 \\
x=3 u+1+v=18 t+6+1+5 t+2=23 t+9 \\
y=x-5+u=23 t+9-5+6 t+2=29 t+6
\end{gathered}
$$

Since we are looking for positive integer solutions, we have $23 t+9>0$, from which

$$
t>-\frac{9}{23}, \text { and } 29 t+6>0, \text { from which } t>-\frac{6}{29}
$$

Thus, the values $t \geq 0$ provide infinitely many solutions.

| $\mathrm{t}$ | 0 | 1 | $2 \ldots$ |
| :---: | :---: | :---: | :--- |
| $\mathrm{x}$ | 9 | 32 | $55 \ldots$ |
| $\mathrm{y}$ | 6 | 35 | $64 \ldots$ |

Remark: We would have reached the goal more quickly if we wrote

$$
x=\frac{23 u+8}{6}=4 u+1+\frac{2-u}{6}=4 u+1+t
$$

from which

$$
\begin{aligned}
& u=2-6 t \\
& x=8-24 t+1+t=9-23 t \\
& y=9-23 t-5+2-6 t=6-29 t
\end{aligned}
$$

Now the values $t \leqq 0$ provide the positive solutions.

Kozma Tibor (Győr, Czuczor G. Gymnasium I. o. t.)