Problem 4

Find all integers $x, y$ and $z$ greater than or equal to 0 such that

$$
2^{x}+9 \times 7^{y}=z^{3}
$$

## $\S$
The statement indicates that $2^{x} \equiv z^{3}(\bmod 9)$. It is easily verified that the only cubes modulo 9 are $-1, 0,$ and $1$, and that $2^{x} \equiv \pm 1(\bmod 9)$ if and only if $x \equiv 0(\bmod 3)$.

Under these conditions, and by setting $t=x / 3$, the equation in the statement can be rewritten as

$$
9 \times 7^{y}=z^{3}-2^{3 t}=\left(z-2^{t}\right)\left(z^{2}+2^{t} z+2^{2 t}\right)
$$

Let $d=\operatorname{PGCD}\left(z-2^{t}, z^{2}+2^{t} z+2^{2 t}\right)$, and then $a=\left(z-2^{t}\right) / d$ and $b=\left(z^{2}+2^{t} z+2^{2 t}\right) / d$. We observe that

$$
0 \equiv z^{2}+2^{t} z+2^{2 t} \equiv 3 \times 2^{2 t} \quad(\bmod d)
$$

and thus $d$ is coprime with 7.

Furthermore, since $b d-(a d)^{2}=3 \times 2^{t} z$, we know that 3 divides either $a d$ and $b d$, or neither of them. Since 9 divides $a b d^{2}$ and $\operatorname{PGCD}(a, b)=1$, we deduce that 3 divides $d$. Conversely, since $d^{2}$ is coprime with 7 and divides $9 \times 7^{y}$, we conclude that $d=3$.

The equality in the statement can then be rewritten as $7^{y}=a b$. Since $a$ and $b$ are coprime, one of them must be equal to 1, and the other to $7^{y}$. Since we have seen that $b>a^{2} d$, it follows that $a=1$ and $b=7^{y}$.

But then $z=2^{t}+3$ and $2^{t} z=\left(b d-(a d)^{2}\right) / 3=7^{y}-3$. Since $7^{y} \equiv \pm 1(\bmod 8)$, we deduce that $2^{t} z \equiv-2$ or $-4(\bmod 8)$, and thus that $t \leqslant 2$. We then check the three possible values of $t$:

$\triangleright$ if $t=0$, our two equations become $z=4$ and $4=7^{y}-3$, so $y=1$;

$\triangleright$ if $t=1$, our two equations become $z=5$ and $10=7^{y}-3$, which is not possible;

$\triangleright$ if $t=2$, our two equations become $z=7$ and $28=7^{y}-3$, which is not possible either.

Therefore, the only possible solution is $(x, y, z)=(0,1,4)$, and it is easily verified that it fits well.