1977 USAMO Problems/Problem 1


Determine all pairs of positive integers $(m,n)$ such that $(1+x^n+x^{2n}+\cdots+x^{mn})$ is divisible by $(1+x+x^2+\cdots+x^{m})$.

Solution 1

Denote the first and larger polynomial to be $f(x)$ and the second one to be $g(x)$. In order for $f(x)$ to be divisible by $g(x)$ they must have the same roots. The roots of $g(x)$ are the (m+1)th roots of unity, except for 1. When plugging into $f(x)$, the root of unity is a root of $f(x)$ if and only if the terms $x^n, x^{2n}, x^{3n}, \cdots x^{mn}$ all represent a different (m+1)th root of unity not equal to 1.

Note that if $\\gcd(m+1,n)=1$, the numbers $n, 2n, 3n, \cdots, mn$ represent a complete set of residues minus 0 modulo $m+1$. However, if $gcd(m+1,n)=a$ not equal to 1, then $\frac{(m+1)(n)}{a}$ is congruent to $0 \pmod {m+1}$ and thus a complete set is not formed. Therefore, $f(x)$ divides $g(x)$ if and only if $\boxed{\\gcd(m+1,n)=1}.$ $\blacksquare$

Solution 2

We could instead consider $f(x)$ modulo $g(x)$. Notice that $x^{m+1} = 1 \pmod {g(x)}$, and thus we can reduce the exponents of $f(x)$ to their equivalent modulo $m+1$. We want the resulting $h(x)$ with degree less than $m+1$ to be equal to $g(x)$ (of degree $m$), which implies that the exponents of $f(x)$ must be all different modulo $m+1$. This can only occur if and only if $gcd(m+1, n) = 1$, and this is our answer, as shown in Solution 1.

See Also

1977 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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