Problem 5

Find all polynomials

P(x)

with integer coefficients such that for all real numbers

s

and

t

, if

P(s)

and

P(t)

are both integers, then

P(st)

is also an integer.

Step 1 of 8: Normalize by a shift and a sign

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P(0)\in\mathbb Z;\quad \deg P=0\Rightarrow P\text{ is constant};\quad \deg P=n\ge1:\ P(x)\longmapsto \pm\big(P(x)-P(0)\big),\quad P(x)=\sum_{i=1}^n a_ix^i,\ a_n>0

Detailed analysis

Since $P$ has integer coefficients, $P(0)$ is automatically an integer. If $P$ is constant, it is already an integer constant and belongs to the final family, so assume $\deg P=n\ge1$ . If $P$ has the stated property then so do $P(x)+k$ for any integer $k$ and $-P(x)$ ; the integrality hypotheses are unchanged. Thus we may replace $P$ by $\pm\big(P(x)-P(0)\big)$ to assume $P(0)=0$ and the leading coefficient $a_n$ of $P(x)=\sum_{i=1}^n a_ix^i$ is positive. It suffices to show $P(x)=x^n$ .