## Task 31/81

We are looking for all Pythagorean number triples $(a ; b ; c)$ for which $(c ; a b ; 4 a-b)$ is also a Pythagorean number triple.
According to the given conditions,

$$
\begin{array}{r}
a^{2}+b^{2}=c^{2} \\
c^{2}+a^{2} b^{2}=16 a^{2}-8 a b+b^{2}
\end{array}
$$

Substituting (1) into (2), we get

$$
a^{2}+b^{2}+a^{2} b^{2}=16 a^{2}-8 a b+b^{2} \rightarrow 15 a^{2}-a^{2} b^{2}-8 a b=0 \rightarrow a=\frac{8 b}{15-b^{2}}
$$

Since $b>3$ implies $a<0$, only $b=3$ is feasible (the numbers 1 and 2 are known not to be part of a Pythagorean triple). Then $a=4$ and $c=5$. Indeed, $(5 ; 4 \cdot 3=12 ; 4 \cdot 4-3=13)$ is a Pythagorean triple. The triple $(4 ; 3 ; 5)$ is thus the only one of the required type.