Problem 6. Calculate the maximum number of distinct real roots that a polynomial $P$ can have, which satisfies the following property: the product of two distinct roots of $P$ is still a root of $P$.
Solution. The answer is 4. The polynomial $x(x-1 / 2)(x-1)(x-2)$, with roots 0, 1/2, 1, and 2, satisfies the given bound. Suppose there was a polynomial with at least 5 distinct roots. It is important to note that a polynomial has a finite number of roots, which allows us to take maximums and minimums.

If a polynomial has 5 roots, at least 4 would be non-zero, and we distinguish two cases.

1. If -1 is a root, there is at least one root (in fact, at least two) with a non-zero absolute value different from 1. Let $A$ be a root with the largest absolute value among all roots and assume that $|A|>1$. Since -1 and $A$ are roots, $-A$ is a root; but then $-A \cdot A=-A^{2}$ would also be a root and $\left|A^{2}\right|>|A|$, which would contradict the maximality of the absolute value of $A$.

If there are no roots with an absolute value $>1$, take $A$ as one of the roots with the smallest absolute value among all non-zero roots. Then $|A|<1$, and since -1 and $A$ are roots, $-A$ is a root and so is $-A \cdot A=-A^{2}$. But then, $\left|A^{2}\right|<|A|$, which would contradict the minimality of the absolute value of $A$ among the non-zero roots.

2. If -1 is not a root, there are at least three non-zero roots with a modulus different from 1. There are then two roots $A, B$ with the two largest absolute values (greater than 1) or two roots with the two smallest absolute values $A, B$ (less than 1 and non-zero). In either case, multiplying $A \cdot B$ leads to a contradiction with the maximality or minimality.