Find all positive integers $a$ such that $a^{2}+2a$ is a perfect square.
There are two cases. If $a>0$, we notice that $a^{2}<a^{2}+2 a<$ $a^{2}+2 a+1=(a+1)^{2}$. Therefore, $a^{2}+2 a$ cannot be a perfect square. If $a=0$, we find $a^{2}+2 a=0$ which is a perfect square. This is therefore the only one.

## 2 Arithmetic Course

This session discussed congruences. The main points covered during the session are addressed in the course on the Animath website, in chapter 3. I will not repeat the entire course, but just provide a summary, referring to the course for more details. The address of this course is at http ://www.animath.fr/spip.php ?article255.

I started by defining congruences (definition 3.1.1), comparing congruences modulo an integer 12 with the hours written on a 12-hour clock, and then by analogy comparing congruences modulo any integer $n$ with the hours written around a hypothetical clock with $n$ hours.

I then emphasized the last property of the course: one can "add" and "multiply" modulo. Intuitively, the property of addition can be visualized by noting that advancing 15 hours from a given position on a 12-hour clock is the same as advancing 3 hours. To visualize the property of multiplication, an idea is to look modulo 10: a number is congruent to its last digit modulo 10, and we know well that to find the last digit of the product of two numbers, it suffices to examine the product of the last two digits.

I then moved on to section 3.2, on divisibility criteria. If an integer is written as $\overline{\text { abcde }}$ in base 10, then it is worth $\mathrm{a} \cdot 10^{4}+\mathrm{b} \cdot 10^{3}+\mathrm{c} \cdot 10^{2}+\mathrm{d} \cdot 10+\mathrm{e}$. If we know the values taken by the sequence $(10)^{n}$ modulo $k$, then we will have a quick method to find the value of $\overline{\mathrm{abcde}}$ modulo $k$.

For example, with $k=7$, we verify that $10 \equiv 3[7], 10^{2} \equiv 2[7], 10^{3} \equiv-1[7]$, and $10^{4} \equiv-3[7]$. Thus, we will have

$$
\begin{aligned}
\overline{\mathrm{abcde}} & \equiv a \cdot(-3)+b \cdot(-1)+c \cdot 2+d \cdot 3+e[7] \\
& \equiv-3 a-b+2 c+3 d+e[7]
\end{aligned}
$$

A concrete example: $\overline{17255} \equiv-3-7+4+15+5[7]$, and therefore it is divisible by 7. We can create as many such criteria as we want.

We see from this example the great strength of the notion of congruence: it allows replacing calculations on large numbers with calculations on much smaller numbers, which are therefore easier. A spectacular example is given by the following exercise. It uses a classic trick, due to the fact that 10 is congruent to 1 modulo 9: a number is always congruent to the sum of its digits modulo 9.