2024 SSMO Team Round Problems/Problem 8

Problem

Three integers $0\le a\le b\le c< 229$ satisfy the congruence $n^3 \equiv 1 \pmod{229}.$ Given that $71^2 - 3$ and $107^2 + 1$ are both multiples of $229,$ find the value of $b.$

Solution

Note that $n^3\equiv1\pmod{229}\implies (n^3-1)\equiv0\pmod{229}.$ Consider the complex third roots of unity $1$ and $\frac{-1\pm i\sqrt{3}}{2}.$ They can be defined based on $\frac12,\sqrt{-1},$ and $\sqrt3.$ From the given multiples of $229,$ we can $\sqrt{3}\equiv71\pmod{229}$ and $i\equiv107\pmod{229}$ since they function like square roots. In addition, $229$ is odd so $2^{-1}$ exists. We have $i\sqrt{3}\equiv107\cdot71\equiv269\pmod{229}.$ Thus, $\frac{-1+i\sqrt{3}}{2} \equiv 134\pmod{229}$ and $\frac{-1-i\sqrt{3}}{2}\equiv94\pmod{229}.$ So, $a = 1,b = \boxed{94},$ and $c = 134.$

~SMO_Team

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2024 SSMO Team Round Problems/Problem 8

Problem

Solution