## Task 14/65

What conditions must $a$ and $b$ satisfy ($a ; n>0$; integers), for $m=\sqrt{a^{n}+a^{n+1}}$ to be a natural number?
It is

$$
m=\sqrt{a^{n}+a^{n+1}}=\sqrt{a^{n}(a+1)}
$$

We now distinguish two cases.

1) Let $n=2 k$ with $k=1 ; 2 ; 3 ; \ldots$. Then we have

$$
\left.m=\sqrt{a^{2 k}(a+1)}\right) a^{k} \sqrt{a+1}
$$

For $m$ to be a natural number, $a+1$ must be a perfect square: $a+1=b^{2 r}$ with $b ; r>0$, integers. This implies $a=b^{2 r}-1$. If $n$ is even, then $a$ must be one less than a power with an even exponent.

2) Let $n=2 k-1$ with $k=1 ; 2 ; 3 ; \ldots$ Then we have

$$
\left.m=\sqrt{a^{2 k-1}(a+1)}\right) a^{k-1} \sqrt{a(a+1)}
$$

For $m$ to be a natural number, $a(a+1)$ must be a perfect square. This is not possible, however, because $a$ and $a+1$ are coprime, so their product can only be a perfect square if each factor is a perfect square, which would imply $a=0$ in contradiction to the assumption $a>0$. The conditions are thus:

1.) $n$ must be an even number and 2.) $a$ must be one less than a power with an even exponent.