Find all pairs of natural numbers $(x, y)$ such that: $7^{x}-3 \times 2^{y}=1$.

## Solutions
On gains by writing the equation differently: $7^{x}-1=3 \times 2^{y}$. Indeed, this allows us to factorize $(7-1)$ in the left-hand side, which gives: $(7-1)\left(7^{x-1}+7^{x-2}+\ldots+7+1\right)=3 \times 2^{y}$, or, by simplifying by 6: $7^{x-1}+7^{x-2}+\ldots+7+1=2^{y-1}$. We see in the initial equation that $x$ cannot be equal to 0. However, it can be equal to 1, in which case $y=1$. If $y>1$,
$2^{y-1}$ is even, which implies that the left-hand side is even, hence the sum of an even number of terms (all odd), so $x$ is even. We can therefore group the terms two by two and factorize: $(7+1)\left(7^{x-2}+7^{x-4}+\ldots+7^{2}+1\right)=2^{y-1}$, or, by simplifying by 8: $7^{x-2}+7^{x-4}+\ldots+7^{2}+1=2^{y-4}$. We see a second solution: $y=4, x=2$. But for $y>4$, $2^{y-4}$ is even, so the first member must again contain an even number of terms, which can be grouped two by two by factoring: $\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\ldots+7^{4}+1\right)=2^{y-4}$. But $7^{2}+1=50$ cannot divide a power of 2, hence the contradiction: there is no solution for $y>4$, and the only solutions to the equation are $(1,1)$ and $(2,4)$.

## 2 PM: Roger Mansuy

## Factorization of the Rectangle

## Proposition

Let $a, b, c$ and $d$ be integers. Then,

$$
a b+b c+c d+d a=(a+c)(b+d)
$$

The term "rectangle" is justified by the following graphical explanation:

![](https://cdn.mathpix.com/cropped/2024_05_10_7286ba4d154441c7d078g-71.jpg?height=372&width=586&top_left_y=1294&top_left_x=737)

## Corollary

Let $x$ and $y$ be integers. Then

$$
\begin{aligned}
& x y+x+y=(x+1)(y+1)-1 \\
& x y-x-y=(x-1)(y-1)-1
\end{aligned}
$$