IMO 1974 · Problem 6

Let \(P\) be a non-constant polynomial with integer coefficients. If \(n(P)\) is the number of distinct integers \(k\) such that \((P(k))^2 = 1\), prove that \(n(P) - \deg(P) \leq 2\), where \(\deg(P)\) denotes the degree of the polynomial \(P\).

Official text of the problem. Official problem paper (PDF)

Solution

Put \(d=\deg(P)\geq 1\). The integers counted by \(n(P)\) are roots of the nonzero polynomial \(P(x)^2-1\), so there are finitely many. If there are none, the desired inequality is immediate.

Otherwise, let \(m\) be the smallest such integer, and put \(s=P(m)\in\{1,-1\}\). Divide the counted integers into those satisfying \(P(k)=s\) and those satisfying \(P(k)=-s\).

The first group has at most \(d\) members, because its members are distinct roots of the nonzero polynomial \(P(x)-s\), which has degree \(d\).

For the second group, recall that for distinct integers \(u,v\), \[ u-v\mid P(u)-P(v). \] Indeed, \(u-v\) divides \(u^j-v^j\) for every positive integer \(j\), and the coefficients of \(P\) are integers.

Thus, if \(P(k)=-s\), then \[ k-m\mid P(k)-P(m)=-2s. \] By the minimality of \(m\), and because \(P(k)\neq P(m)\), we have \(k>m\). Consequently, \(k-m\) is a positive divisor of \(2\), so \[ k\in\{m+1,m+2\}. \] The second group therefore has at most two members. Combining the two bounds gives \[ n(P)\leq d+2, \] as required. \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).