Find all pairs of positive integers $(x, y)$ such that $7^{x}-3\cdot2^{y}=1$
The equation has a first, quite obvious solution: $x=y=1$, clearly the only solution for $y=1$. The statement does not exclude the possibility that $y$ is zero, but for $y=0$, it is clear that we cannot have $7^{x}=4$. Now suppose $y \geq 2$: $3.2^{y}$ is divisible by $3.2^{2}=12$, so a necessary condition for the equation to hold is that $7^{x} \equiv 1(\bmod 12)$. Modulo $12,7^{x}$ takes only two values: 7 when $x$ is odd, 1 when $x$ is even. Therefore, for $y \geq 2$, any solution $(x, y)$ must necessarily satisfy: $x$ even. If $x=2k$, the equation can be written as: $3.2^{y}=7^{2k}-1=\left(7^{k}+1\right)\left(7^{k}-1\right)$. The factor 3 must necessarily divide $7^{k}-1$ because $7 \equiv 1(\bmod 3)$, so for all $k, 7^{k} \equiv 1(\bmod 3)$. Therefore, $7^{k}-1=3 \cdot 2^{z}$ and $7^{k}+1=2^{y-z}$. A priori, we do not know $z$, but for $7^{k}+1=2^{y-z}>3.2^{z}=7^{k}-1$, it is necessary that $y-z>z$. Consequently, $2^{y-z}-3.2^{z}=\left(7^{k}+1\right)-\left(7^{k}-1\right)=2$ must be divisible by $2^{z}$, which implies $z=1$ or $z=0$, so $7^{k}-1=6$ or $7^{k}-1=3$. Only the first possibility provides a solution: $k=1$ hence $x=2$ and $y=4$. The Diophantine equation therefore has two solutions in total: $(1,1)$ and $(2,4)$.

## 2 TD

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