Let $a$ be a strictly positive integer. Suppose that $4\left(a^{n}+1\right)$ is the cube of an integer for all positive integers $n$. Find $a$.

Soit $a$ un entier strictement positif. On suppose que $4\left(a^{n}+1\right)$ est le cube d'un entier pour tout entier positif $n$. Trouver $a$.

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It is clear that $a=1$ works. Let's show that it is the only one. Suppose therefore $a>1$. Choose $n=2 m$ and note that $a^{2}+1$ is not a power of 2 since it is congruent to 1 or 2 modulo 4. Let $p$ be an odd prime such that $p$ divides $a^{2}+1$. Then, according to LTE:

$$
v_{p}\left(4\left(a^{n}+1\right)\right)=v_{p}\left(a^{2}+1\right)+v_{p}(m)
$$

We choose $m$ so that the latter term is congruent to 1 modulo 3. Then $4\left(a^{n}+1\right)$ cannot be a cube, contradiction.