## Task 11/82

Determine all ordered quadruples $(k ; l ; m ; n)$ of positive integers $k, l, m, n$ for which the following equation holds: $k^{2 l}-4=m^{n}$.
The equation has infinitely many trivial solutions. It is certainly satisfied by the two-parameter solution set

$$
\left(k ; l ; m=k^{2 l}-4 ; n=1\right)
$$

with $k, l \in \mathbb{N}; k>2$. The question of the existence of non-trivial solutions is quickly resolved if $m$ is a prime number. Then we have

$$
k^{2 l}-4=\left(k^{l}-2\right)\left(k^{l}+2\right)=m^{n}=m^{\alpha} m^{n-\alpha}
$$

with $\alpha \in \mathbb{N}, m^{\alpha} \neq m^{n-\alpha}$ because $k^{l}-2<k^{l}+2$. Without loss of generality, let $m^{\alpha}<m^{n-\alpha}$, so $\alpha<n-\alpha, 2 \alpha<n$. Then,

$$
\frac{k^{l}+2}{k^{l}-2}=1+\frac{4}{k^{l}-2}=\frac{m^{n-\alpha}}{m^{\alpha}}=m^{n-2 \alpha} \in \mathbb{N}
$$

Consequently, $k^{l}=3$ or $k^{l}=4$ or $k^{l}=6$. Only $k^{l}=6$ provides a non-trivial solution: $(k ; l ; m ; n)=$ $(6 ; 1 ; 2 ; 5)$.

If $m$ is not a prime number, then $n$ is certainly odd; because from $k^{2 l}-4=m^{2 r}$ with $r \in \mathbb{N}$ it follows that two square numbers $k^{2 l}$ and $m^{2 r}$ differ by exactly 4, which is known to be impossible. Let now $n=2 r+1$ with $r \in \mathbb{N}$. Then,

$$
k^{2 l}=m^{2 r+1}+4=m^{2 r}\left(m+\frac{4}{m^{2 r}}\right)
$$

must be a square number. For the expression in parentheses to be an integer, $m^{2 r}$ can only take the values 1, 2, or 4. Therefore, $m=1$ and $r$ arbitrary, or $m$ arbitrary and $r=0$, or $m=2$ and $r=1$. For $m=1$ and $m=2$, no square numbers result. For $r=0$, $n=1$, and thus one obtains the already mentioned trivial solutions. Further non-trivial solutions follow from trivial ones if $k=s^{2}$ with $s \in \mathbb{N}$.