The author of the problem is George STOICA from Canada.
Although proposed for 12th grade (the last grade before college), it only requires elementary knowledge of quadratic equations and some common sense knowledge of natural numbers.
Here we denote $\mathbb{N}=\{0,\;1,\;2,\dots \}$ be the set of natural numbers, $\mathbb{Q}$ the set of rational numbers. We state the following, almost obvious, statement as a
Lemma. Let $m\in\mathbb{N}$ such that $\sqrt{m}\in\mathbb{Q}$. Then $\sqrt{m}\in\mathbb{N}$.
Proof CiP $\sqrt{m}$ admits the representation $\sqrt{m}=\frac{q}{r},\;\;q,\;r\in\mathbb{N},\;r\neq0$, where the greatest common divisor of the numbers $q$ and $r$, $(q,r)=1$. By the Fundamental Theorem of Arithmetic, we have the (unique) writings
$m=p_1^{e_1}\cdot p_2^{e_2}\cdot \dots \cdot p_k^{e_k},\;\;q=p_1^{f_1}\cdot p_2^{f_2}\cdot \dots \cdot p_k^{f_k},\;\;r=p_1^{g_1}\cdot p_2^{g_2}\cdot \dots \cdot p_k^{g_k}\;\;\;\;k\geqslant 1 \tag{1}$
where $p_1<p_2<\dots <p_k$ are primes and $e_{1-k},\;f_{1-k},\;g_{1-k}$ are nonnegative integers. In (1), some $e_i,\;f_i,\;g_i$ can be zero and the condition $(q,r)=1$ requires that, for any index $i$, one of $f_i\; or\; g_i$ be zero.
From the equality $r^2\cdot m=q^2$ and the uniqueness of the representation as a product of powers of prime numbers, it follows
$2\cdot g_i+e_i=2\cdot f_i$
so all the exponents $e_i$ are even numbers and then $\sqrt{m}=p_1^{e_1/2}\cdot p_2^{e_2/2}\cdot \dots \cdot p_k^{e_k/2} \in\mathbb{N}$.
$\square$Lemma
"Let $P$ be a polynomial with integer coefficients and $a,\;b\in\mathbb{Z}$ such
that $P(a)\cdot P(b)=-(a-b)^2$. Show that $P(a)=-P(b)=\pm(a-b)$."
ANSWER CiP
An EXAMPLE of such a polynomial is $P(X)=2\cdot X-a-b$
Solution (as it appears in GMB 1/2026, page 38, adapted by CiP)
From the algebraic identity
$\frac{a^k-b^k}{a-b}=a^{k-1}+a^{k-2}b+\dots+ab^{k-2}+b^{k-1},\;\;\forall k\geqslant 1$
it follows that for any polynomial $P$ , the number $\alpha=\frac{P(a)-P(b)}{a-b}\in \mathbb{Z}$.
For rational numbers
$x_1=\frac{P(a}{a-b},\;\;\;x_2=-\frac{P(b}{a-b} \tag{2}$
we have $x_1+x_2=\frac{P(a)-P(b)}{a-b}=\alpha,\;\;x_1\cdot x_2=-\frac{P(a)\cdot P(b)}{(a-b)^2}=1$ so the numbers (2) are the roots of the equation with integer coefficients
$x^2-\alpha \cdot x+1=0. \tag{3}$
Then the discriminant of the equation (3) is a rational number.
$\blacksquare$
<to be continue>
