Hypothesis of recurrence: at rank $n, 1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}} \leq 2-\frac{1}{n}$. Initialization: for $n=1,1 \leq 2-\frac{1}{1}$, so the relation is clearly true at rank 1.

Calculation: the relation $1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}} \leq 2-\frac{1}{n}$ being assumed true at rank $n$ (hypothesis of recurrence), to transform this relation into the same relation at rank $n+1$, it is necessary to add $\frac{1}{(n+1)^{2}}$ on the left and on the right: $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$ which is clearly larger. Therefore, the inequality remains true, and by recurrence it is true for all $n \geq 1$. We deduce that for all $n \geq 1$, $1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}} \leq 2-\frac{1}{n}<2$.

## 2 Group A: Algebra

## 1 Friday morning, 18: Mathieu Barré

## Algebraic manipulations

Calculating with letters requires knowing a few identities. In what follows, unless otherwise stated, the letters used represent real numbers.

## Properties to remember

1. Distributivity: $k(a+b)=k a+k b$.
2. Rectangle factorization: $a b+b c+c d+d a=(a+c)(b+d)$.
3. If $a$ is positive, $\left(a^{b}\right)^{c}=\left(a^{c}\right)^{b}=a^{b c}$.
4. $(a+b)^{2}=a^{2}+2 a b+b^{2}$.
5. $(a-b)^{2}=a^{2}-2 a b+b^{2}$.
6. $a^{2}-b^{2}=(a+b)(a-b)$.
7. $(a+b)^{3}=a^{3}+3 a^{2} b+3 b^{2} a+b^{3}$.
8. $(a-b)^{3}=a^{3}-3 a^{2} b+3 b^{2} a-b^{3}$.
9. $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$.
10. More generally, if $n$ is a natural number,

$$a^{n}-b^{n}=(a-b)\left(a^{n-1}+a^{n-2} b+\cdots+a b^{n-2}+b^{n-1}\right)$$

In particular, taking $b=1$, we get

$$a^{n}-1=(a-1)\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)$$

11. Binomial theorem: if $n$ is an integer, then

$$(a+b)^{n}=\sum_{i=0}^{n}\binom{n}{i} a^{n-i} b^{i}$$

It is always important to keep these properties in mind in order to recognize them in exercises, generally to obtain a factored form.

## Polynomial division

Another idea for factoring algebraic expressions is to generalize the notion of division practiced since elementary school. For example, if we want to divide $a^{3}-1$ by $a-1$, we can imagine that $a^{3}-1$ and $a-1$ are integers and perform the division as an elementary school student would. We first try to eliminate the term in $a^{3}$ by asking: "By what must we multiply $a-1$ to get close to $a^{3}$?". The answer is $a^{2}$ and $a^{2}(a-1)=a^{3}-a^{2}$. We then subtract $a^{3}-a^{2}$ from $a^{3}-1$: the $a^{3}$ terms cancel out (it's designed to do so!) and we are left with $a^{3}-1-\left(a^{3}-a^{2}\right)$, which is $a^{2}-1$. We then repeat this procedure with $a^{2}-1$. The different steps are summarized in the division below:

$$\text{a 3 -1 a-1 - (a 3 -a 2 ) a 2 +a+1 a 2 -1 - (a 2 -a ) a-1 -(a-1) 0}$$

We have thus obtained: $a^{3}-1=(a-1)\left(a^{2}+a+1\right)$, which is also a consequence of property 9 for $b=1$.

Let's continue to practice this principle with another example: factorize $a^{4}+4$ by $a^{2}-2 a+2$. Proceeding exactly as before, we get successively:

$$\text{a 4 +4 a 2 -2 a+2 - (a 4 -2 a 3 +2 a 2 ) a 2 +2 a+2 2 a 3 -2 a 2 +4 .a 2 +2 a ) - (2 a 3 -4 a 2 +4 a ) 2 a 2 -4 a+4 - (2 a 2 -4 a+4 )}$$

Thus, the final result is $a^{4}+4=\left(a^{2}-2 a+2\right)\left(a^{2}+2 a+2\right)$, known as Sophie Germain's identity.

We now have a method (or even an algorithm) for performing polynomial divisions. However, in the two previous examples, we were lucky because the remainders of the divisions were zero, which is not always the case. How can we guess that a polynomial will be able to be factored by another? The following theorem provides an answer:

Theorem 3. Let $P(X)$ be a polynomial and $\alpha$ a root of $P$ (i.e., a real number that satisfies $P(\alpha)=0$). The polynomial $P(X)$ can then be factored by $X-\alpha$.

Corollary 4. A polynomial of degree $n$ has at most $n$ distinct roots.

Proof. The proof of this result (as well as many others that will interest the curious reader) can be found in Igor Kortchemski's course on polynomials, available here.

We better understand why the polynomial $Q(a)=a^{3}-1$ can be factored by $a-1$: it's because 1 is a root of $Q$, since $Q(1)=1^{3}-1=0$. The corollary is also very useful. Indeed, if we have a polynomial of degree $n$ for which we know exactly $n$ roots $\alpha_{1}, \ldots, \alpha_{n}$, it will necessarily be written in the form $\left(X-\alpha_{1}\right) \cdots\left(X-\alpha_{n}\right)$, up to multiplication by a constant.

Example 5. We already know how to factor the polynomial $P(X)=X^{2}-2$ using the remarkable identity 6, but let's try to achieve this using our new result. We notice that $\sqrt{2}$ and $-\sqrt{2}$ are two roots of $P$, which is of degree 2: we have thus found all its roots. Since the leading coefficient of $P$ is 1, we automatically have $P(X)=(X-\sqrt{2})(X+\sqrt{2})$.

Remark 6. Showing