266. $43 x+41=39 y+33=35 z+25=31 u+17$.

Problem from a German algebraic manuscript (No. 14908 of the Munich collection).
266. Given: $43 x+8=39 y ; 39 y+8=35 z ; 35 z+8=31 u$ or: $43 x+16=35 z=31 u-8$, and therefore, $43 x+24=31 u$.

$$
\begin{gathered}
u=\frac{43 x-24}{31}=x+1+\frac{12 x-7}{31}=x+1+t \\
12 x-7=31 t \\
x=\frac{31 t+7}{12}=2 t+\frac{7 t+7}{12}=2 t+7 t_{1}, \text { where } t_{1}=\frac{t+1}{12} \\
t=12 t_{1}-1 ; \quad x=31 t_{1}-2 ; u=43 t_{1}-2 ; \quad y=\frac{31 \cdot 43 t_{1}}{39}-2
\end{gathered}
$$

and

$$
z=\frac{31 \cdot 43 t_{1}}{35}-2
$$

For $z$ and $y$ to be integers, it is necessary that $t_{1}=39 \cdot 35$ (at least). Then

$$
\begin{aligned}
& x=31 \cdot 39 \cdot 35-2 \\
& y=31 \cdot 35 \cdot 43-2 \\
& z=31 \cdot 39 \cdot 43-2 \\
& u=35 \cdot 39 \cdot 43-2
\end{aligned}
$$