IMC 2015 · Day 1 · P4

We show that such integers $m_1, \dots, m_{15}$ do not exist.

Suppose that (1) is satisfied by some integers $m_1, \dots, m_{15}$. Then the argument of the complex number $z_1 = 1 + 16i$ coincides with the argument of the complex number $$z_2 = (1 + i)^{m_1} (1 + 2i)^{m_2} (1 + 3i)^{m_3} \cdots\cdots (1 + 15i)^{m_{15}}.$$ Therefore the ratio $R = z_2 / z_1$ is real (and not zero). As $\operatorname{Re} z_1 = 1$ and $\operatorname{Re} z_2$ is an integer, $R$ is a nonzero integer.

By considering the squares of the absolute values of $z_1$ and $z_2$, we get $$(1 + 16^2) R^2 = \prod_{k=1}^{15} (1 + k^2)^{m_k}.$$ Notice that $p = 1 + 16^2 = 257$ is a prime (the fourth Fermat prime), which yields an easy contradiction through $p$-adic valuations: all prime factors in the right hand side are strictly below $p$ (as $k < 16$ implies $1 + k^2 < p$). On the other hand, in the left hand side the prime $p$ occurs with an odd exponent.