International Mathematics Competition for University Students

July 27 - Aug 2 2015, Blagoevgrad, Bulgaria


Day 1
    Problem 1
    Problem 2
    Problem 3
    Problem 4
    Problem 5

Day 2
    Problem 6
    Problem 7
    Problem 8
    Problem 9
    Problem 10


    Day 1 questions
    Day 1 solutions
    Day 2 questions
    Day 2 solutions

Official IMC site

Problem 4

4. Determine whether or not there exist 15 integers $m_1,\ldots,m_{15}$ such that~ $$\displaystyle \sum_{k=1}^{15}\,m_k\cdot\arctan(k) = \arctan(16). \qquad\qquad(1)$$

Proposed by Gerhard Woeginger, Eindhoven University of Technology

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

Suppose that (1) is satisfied by some integers $m_1,\ldots,m_{15}$. 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 $\re z_1=1$ and $\re z_2$ 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.