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 5 of 8: Contradiction rules out a proper factor

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P(r)=p,\qquad |P(r)|\le\sum_{i=1}^n|a_i|\,|r|^i\le\sum_{i=1}^n|a_i|<p\quad\text{(contradiction)}

Detailed analysis

Since $|r|\le1$ and $P(x)=\sum_{i=1}^n a_ix^i$ , $|P(r)|\le\sum_{i=1}^n|a_i||r|^i\le\sum_{i=1}^n|a_i|<p$ by the choice of $p$ . But $P(r)=p$ from step 4, so $p<p$ , a contradiction. Hence $f$ cannot be a proper factor: $f$ is a scalar multiple of $P(x)-p$ itself, so $\big(P(x)-p\big)\mid\big(P(2x)-P(2t)\big)$ .