Conjecture.Let $P(x),Q(x) \in \mathbb{R}[x]$ be two monic polynomials with non-negative coefficients. If $R(x)=P(x)Q(x)$ is $0,1$ polynomial (coefficients only from $\{0,1\}$), then $P(x)$ and $Q(x)$ are also $0,1$ polynomials.

Is this conjecture true or is there a counter-example? If non-monics were allowed, we would have $x^2=2x\cdot (x/2)$, and with negative coefficients we could find for example $x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$. However with these two conditions the conjecture seems to hold for $\deg R \leq 12$ which I did verify by Maple (although a numerical errors in Maple factorization might have caused some false negatives).

This question has been asked on MSE in more than one week ago: https://math.stackexchange.com/questions/3325163/the-coefficients-of-a-product-of-monic-polynomials-are-0-and-1-if-the-polyn, but no solution has been found.

**Edit (probability theory reformulation):** An interesting alternative reformulation provided in comments by Gro-Tsen/Rémy Oudompheng: Assume $X,Y$ are independent random variables supported on a finite subset of the integers, and assume $Z=X+Y$ is uniformly distributed on its support: is it necessarily the case that $X$ and $Y$ are themselves uniformly distributed on their support? (Hence the [pr.probability] and [probability-distributions] tags).

**Edit (further verification):** As mentioned in the comments, Max Alekseyev has extended the verification to all degrees up to and including $26$, which was further improved up to degree $32$ by Peter Mueller using Groebner basis calculation (an approach without numerical issues).

55more comments