2021 Mediterranean MO #1

Determine the smallest positive integer $M$ with the following property: For every choice of integers $a,b,c$, there exists a polynomial $P(x)$ with integer coefficients so that $P(1)=aM$ and $P(2)=bM$ and $P(4)=cM$.




Taking $a=1, b=2, c=3$, we see that $2\mid M(c-b)$ and $3\mid M(c-a)$ must hold true, which implies that $6\mid M$. Now, we show that $M=6$ is a valid possibility. As a construction, consider the polynomial$$P_{abc}=(2a+c-3b)x^2+(15b-12a-3c)x+(16a+2c-12b).$$In this case, we have $M=6$, as desired. $\square$


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