(IMO 1975, Problem 6)

Let $n$ be a positive integer. Find all polynomials $f$ in two variables such that:

1. for all real numbers $t, x, y: f(t x, t y)=t^{n} f(x, y)$;
2. for all real numbers $a, b, c: f(a+b, c)+f(b+c, a)+f(c+a, b)=0$;
3. $f(1,0)=1$.
The first condition indicates that we are looking for a homogeneous polynomial of degree $n$ in two variables.

We can apply the second condition for $a=b=c$, which gives us that $f(2 y, y)=0$ for all real $y$. We deduce that $X-2 Y$ divides $f$.

We know that the irreducible factors of $f$ are of the form $X-\lambda Y$ with $\lambda \in \mathbb{C}$ and the coefficient in front of $X^{n}$ is 1 (since $f(1,0)=1$). We write $f(X, Y)=(X-2 Y) g(X, Y)$ with $g$ a homogeneous polynomial of degree $n-1$ and $g(1,0)=1$. We will show that for all real $x, y$ such that $x+y=1$ we have $g(x, y)=1$, which implies $g(X, Y)=(X+Y)^{n-1}$. Indeed, for all $x, y$ such that $x+y \neq 0$ we have $g(x, y)=(x+y)^{n-1} g\left(\frac{x}{x+y}, \frac{y}{x+y}\right)=(x+y)^{n-1}$ and then conclude by continuity.

We want to show that $x+y=1$ implies $f(x, y)=x-2 y$, i.e., $f(x, 1-x)=3 x-2$ for all $x$. We perform the following change of variable: we set $z=3 x-2$ and $f(x, 1-x)=P(z)$, where $P$ is a polynomial in one variable. By applying the second condition with $a=2 x-1$ and $b=c=1-x$, we have that for all $x, 2 f(x, 1-x)+f(2-2 x, 2 x-1)=0$, hence $2 P(z)+P(-2 z)=0$ for all $z$. Therefore, $P(z)=-\frac{1}{2} P(-2 z)$ and by iterating: $P(z)=\left(-\frac{1}{2}\right)^{k} P\left((-2)^{k} z\right)$ for all integer $k$. By letting $k$ tend to infinity, we have that $P$ is of degree 1 and has a zero constant term. Moreover, we know that $P(1)=1$ (third condition). We deduce that $P(z)=z$, which is what we wanted. Finally, the only solution is $f(X, Y)=(X-2 Y)(X+Y)^{n-1}$.

How to think about looking at $x+y=1$? Perhaps by trying to exploit the second condition by making $f(\lambda c, c)$ appear where $X-\lambda Y$ is a potential factor. For example, by setting $a+b=\lambda c$ and $a=c$, we get $2 f(\lambda c, c)+f(2 c,(\lambda-1) c)=0$ so $f(\lambda c, c)=0$ for all $c$ if and only if $f\left(\frac{2}{\lambda-1} c, c\right)=0$ for all $c$ (where $\lambda=1$ does not work according to the third condition). By looking for which $\lambda$ the two values $\lambda$ and $\frac{2}{\lambda-1}$ are equal, we find $\lambda=2$, which we already had, or $\lambda=-1$, which corresponds to the factor $(X+Y)^{n-1}$. We can also show that $f(0,1)=-2$, so the coefficient in front of $Y^{n-1}$ in $g$ is 1 or -1.

## 3 Friday morning: Polynomials, Antoine Séré

## Parity

Let $P$ have integer coefficients, such that $P(0)$ and $P(1)$ are odd. Show that $P$ has no integer roots.

Solution: pass the problem modulo 2.

## Prime Numbers

Prove that there is no non-constant polynomial $P$ such that $P(0), P(1), P(2), \ldots$ are all prime.

Solution: look at the $P(k P(0))$. They are equal to $P(0)$ because they are prime and divisible by $P(0)$. Therefore, $P$ is constant, which is impossible.

## Images of Squares

Let $P$ be a polynomial of degree $n$. Suppose that $P(0), P(1), P(4), \ldots, P\left(n^{2}\right)$ are integers. Show that for all integers $i, P\left(i^{2}\right)$ is an integer.

Solution: introduce $Q(X)=P\left(X^{2}\right)$, and use Lagrange or Newton interpolation polynomials.

## Fibonacci

Let $P$ be of degree at most 990, such that for all integers $k$ between 992 and 1982, $P(k)=F_{k}$ where $F$ is the Fibonacci sequence. Give a simple expression for $P(1983)$.

Solution: use Lagrange or Newton interpolation polynomials.

## Geometric Interpretation?

Let $P(X)=X^{4}-p X^{3}+q X^{2}-r X+s$. We know that there exists a triangle $ABC$ such that $\tan (A), \tan (B), \tan (C)$ are three of the four roots of $P$. Express the fourth root in terms of $p, q, r, s$ only.

Solution: recall or find the relation: $a=\tan (A)+\tan (B)+\tan (C)=\tan (A) \tan (B) \tan (C)$, true for any triangle $ABC$.

Write $X^{4}-p X^{3}+q X^{2}-r X+s=(X-R)\left(X^{3}-a X^{2}+b X-a\right)=X^{4}-(R+a) X^{3}+(R a+b) X^{2}-(R b+a) X+R a$.

By identifying the coefficients, we find a system of equations. By manipulating these equations, we find:

$$
R=\frac{r-p}{q-s-1}
$$

## Brute Force

Find the coefficient of $x^{2}$ in $P_{n}$ defined by:

$P_{0}=X$

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

Solution: find by induction the form of the leading coefficient of $P_{n}$, then that of its coefficient in $X$, and then that of the coefficient in $X^{2}$.

## 4 Saturday morning: Inequalities, Guillaume Conchon-Kerjan

## Course on Inequalities

Whether at the IMO or in other international competitions, inequalities have been on the decline for several years. However, it is important to know the standard techniques, which could be useful in arithmetic or combinatorial problems. We thus reviewed a number of classical inequalities (Schur, Muirhead, arithmetic-geometric inequality), and addressed the so-called "triangle method" for homogeneous inequalities in three variables. We concluded with a slightly more refined property than the usual inequalities, known as the "SOS theorem," which can be useful in some tricky exercises.

The course is based on that of Matthieu Piquerez at the 2015 Valbonne Olympiad stage, and the last part can be found in Evan Chen's "SOS theorem" paper.

For a general course on inequalities and a substantial set of exercises, one can consult Evan Chen's "Inequalities" article as well as those of Thomas Mildorf and Kiran Kedlaya on the corresponding page of Art of Problem Solving.

## Exercises