. Find all pairs $(x, y)$ of natural numbers such that:

$$
1+x+x^{2}+x^{3}+x^{4}=y^{2}
$$
. The key idea to approach this exercise is to see in the "two largest terms" of the left-hand side the beginning of a square: $x^{4}+x^{3}$ is the beginning of $\left(x^{2}+\frac{x}{2}\right)^{2}=x^{4}+x^{3}+\frac{x^{2}}{4}$, so that for all $x$ and all $y$ natural integers, $y^{2}=\left(x^{2}+\frac{x}{2}\right)^{2}+\frac{3 x^{2}}{4}+x+1$, hence in particular $y>x^{2}+\frac{x}{2}$. If we prove that $y<x^{2}+\frac{x+1}{2}$, we have $\frac{(x+1)(x-3)}{4}>0$, so $x^{2}+\frac{x+1}{2}>y$, which proves that there is no integer $y$ satisfying the equation. However, if $x=3$, the quadratic is zero, $y=x^{2}+\frac{x+1}{2}=11$ : we have here an integer solution of the equation, $1+3+3^{2}+3^{3}+3^{4}=11^{2}$. For $x<3$, our method does not allow us to conclude, and these few remaining cases must be studied by hand:

(ヨe $x=2,1+2+2^{2}+2^{3}+2^{4}=31$ is not the square of an integer,

( $x=1,1+1+1+1+1=5$ is not the square of an integer either,

(ઉ) $x=0,1+0+0+0+0=1$ is the square of an integer, 1 .

In the end, we have two pairs of solutions: $(0,1)$ and $(3,11)$.