Some solves/partial progress.
(JMO Q6) Let $n \geq 2$ be an integer. Let $P(x_1, x_2, \ldots, x_n)$ be a nonconstant $n$-variable polynomial with real coefficients. Assume that whenever $r_1, r_2, \ldots , r_n$ are real numbers, at least two of which are equal, we have $P(r_1, r_2, \ldots , r_n) = 0$. Prove that $P(x_1, x_2, \ldots, x_n)$ cannot be written as the sum of fewer than $n!$ monomials. (A monomial is a polynomial of the form $cx^{d_1}_1 x^{d_2}_2\ldots x^{d_n}_n$, where $c$ is a nonzero real number and $d_1$, $d_2$, $\ldots$, $d_n$ are nonnegative integers.)
Solution. It’s quite easy to argue that $(x_i-x_j)$ divides $P$, and then we can assume that
$$P = \prod_{i The rest is actually somewhat combinatorics. For $n=2$, there are at least 2 monomials because the following are distinct: For $n=3$: similar idea: and generalize to $n$. Why does this work? Note that the $P$-monomial corresponding to the $Q$-monomial $x_{i_1}^{n-1}x_{i_2}^{n-2}...x_{i_{n-1}}$ in $Q$ will be the monomial of maximal deg in $x_{i_1}$, and among these monomial it will be the monomial of maximal deg in $x_{i_2}$ etc, so they cannot overlap (since another monomial cannot be maximal degree in $x_{i_1}$ unless $x_{i_1}^{n-1}$ divides the $Q$-monomial) Let $p$ be an odd prime. An integer $x$ is called a quadratic non-residue if $p$ does not divide $x-t^2$ for any integer $t$. Denote by $A$ the set of all integers $a$ such that $1\le a I cheated a bit. Here’s the python code: Result: I ended up doing this because I made a mistake on $p=17$ and thought the answer wasn’t 2. But anyway… Ideas: Let $n \ge 2$ be an integer. Let $x_1 \ge x_2 \ge ... \ge x_n$ and $y_1 \ge y_2 \ge ... \ge y_n$ be $2n$ real numbers such that$$0 = x_1 + x_2 + ... + x_n = y_1 + y_2 + ... + y_n $$$$\text{and} \hspace{2mm} 1 =x_1^2 + x_2^2 + ... + x_n^2 = y_1^2 + y_2^2 + ... + y_n^2.$$Prove that$$\sum_{i = 1}^n (x_iy_i - x_iy_{n + i - 1}) \ge \frac{2}{\sqrt{n-1}}.$$
# USAMO 2020 Q3
from functools import reduce
def get_A(p):
return [a for a in range(p) if notqr(a,p) and notqr(4-a,p)]
def notqr(a,p):
A = [a for i in range((p-1)//2)]
return (prod_mod_p(A,p)+1) % p == 0
def prod_mod_p(A,p):
return reduce(lambda x,y: (x * y) % p, A, 1)
ps = [3,5,7,11,13,17,19,23,29]
for p in ps:
A = get_A(p)
prod = prod_mod_p(A,p)
print(f'For p={p}: prod={prod}, A={A}')
For p=3: prod=2, A=[2]
For p=5: prod=2, A=[2]
For p=7: prod=2, A=[5, 6]
For p=11: prod=2, A=[2, 7, 8]
For p=13: prod=2, A=[2, 6, 11]
For p=17: prod=2, A=[7, 10, 11, 14]
For p=19: prod=2, A=[2, 8, 10, 13, 15]
For p=23: prod=2, A=[5, 7, 10, 17, 20, 22]
For p=29: prod=2, A=[2, 12, 14, 15, 18, 19, 21]