Exercise 12. Let $(a_n)$ be a sequence of real numbers. Suppose that $a_0 = 1$ and for all $n \geqslant 1, a_n$ is the smallest strictly positive solution of

$$
\left(a_n - a_{n-1}\right)\left(a_n + a_{n-1} - 2 \sqrt{n}\right) = 2
$$

Find the smallest integer $n$ such that $a_n \geqslant 2022$.
Solution to Exercise 12 To get an idea of the problem, we calculate the first values of $a_{n}$: for $a_{1}$, the equation is

$$
\left(a_{1}-1\right)\left(a_{1}-1\right)=2
$$

so $a_{1}-1=\sqrt{2}$ or $-\sqrt{2}$. Only the first case gives a positive solution $a_{1}=1+\sqrt{2}$. Now let's try to calculate $a_{2}$: the equation becomes

$$
\left(a_{2}-1-\sqrt{2}\right)\left(a_{2}+1-\sqrt{2}\right)=2
$$

which can be expanded to

$$
\left(a_{2}-\sqrt{2}\right)^{2}-1^{2}=2
$$

or $a_{2}-\sqrt{2}= \pm \sqrt{3}$. Only one sign gives a positive solution $a_{2}=\sqrt{2}+\sqrt{3}$.

From these first values, we hypothesize that the sequence $a_{n}$ is given by $a_{n}=\sqrt{n}+\sqrt{n+1}$ and we prove this by induction. It is true for $n=0$. Suppose it is true for $n-1$ and show it for $n$: the equation giving $a_{n}$ is

$$
\left(a_{n}-\sqrt{n-1}-\sqrt{n}\right)\left(a_{n}+\sqrt{n-1}-\sqrt{n}\right)=2
$$

which can be expanded to

$$
\left(a_{n}-\sqrt{n}\right)^{2}-(\sqrt{n-1})^{2}=2
$$

and thus

$$
\left(a_{n}-\sqrt{n}\right)^{2}=n+1
$$

so $a_{n}-\sqrt{n}= \pm \sqrt{n+1}$. Only the positive sign gives a positive solution, so we have $a_{n}=\sqrt{n}+$ $\sqrt{n+1}$, which concludes the induction.

We immediately notice that the sequence $(a_{n})$ is increasing, and consequently, if we find an $n$ such that $a_{n-1}<2022 \leqslant a_{n}$, this is the $n$ we are looking for. Indeed, if we have $0 \leqslant m<n$, then $m \leqslant n-1$ and thus $a_{m} \leqslant a_{n-1}<2022$ as required: $n$ is indeed the smallest index such that $a_{n} \geqslant 2022$.

But now we have

$$
a_{1011^{2}-1}=\sqrt{1011^{2}-1}+1011<2 \cdot 1011=2022<1011+\sqrt{1011^{2}+1}=a_{1011^{2}}
$$

so the $\mathrm{n}$ we are looking for is $\mathrm{n}=1011^{2}$.

Graders' Comments: The problem was well solved by the students who submitted a solution. A vast majority perfectly conjectured $a_{n}=\sqrt{n}+\sqrt{n+1}$ and then proved it by induction. It is worth noting that calculating the first terms to guess an expression is a very good initiative to take in the rough draft as it can be the first step in solving a problem (as was the case here). From there, one could quickly conclude, but it was important to avoid calculation errors and to provide proper justification (arguments of increasing were relevant). These points were missing in some solutions. It is also worth noting that in tests or competitions, there is no calculator, so it is important to get used to performing efficient calculations by hand.