## Task 13/83

We are looking for all triples $(a ; b ; c)$ of positive integers with $c>1$, that satisfy the Diophantine equation $a^{2 c}-b^{2 c}=665$.
Due to

$$
a^{2 c}-b^{2 c}=\left(a^{c}-b^{c}\right)\left(a^{c}+b^{c}\right)=665=1 \cdot 5 \cdot 7 \cdot 19
$$

and because $a^{c}-b^{c}<a^{c}+b^{c}$, only the following possibilities are feasible:

$$
\begin{array}{lcc}
\text { 1. } & a^{c}-b^{c}=1 & a^{c}+b^{c}=5 \cdot 7 \cdot 19=665 \\
\text { 2. } & a^{c}-b^{c}=5 & a^{c}+b^{c}=7 \cdot 19=133 \\
\text { 3. } & a^{c}-b^{c}=7 & a^{c}+b^{c}=5 \cdot 19=95 \\
\text { 4. } & a^{c}-b^{c}=19 & a^{c}+b^{c}=5 \cdot 7=35
\end{array}
$$

The first three systems of equations do not yield solutions in whole numbers. The fourth system results in $a^{c}=27, b^{c}=8$, thus $a=3, b=2, c=3$. Indeed, $3^{6}-2^{6}=729-64=665$. The only triplet is therefore $(3 ; 2 ; 3)$.