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gcd(c^2 + d^2, a^2 + b^2) > 1. when ac+bd is divisible by a^2 +b^2

Source: 2015 Saudi Arabia IMO TST IV p1

July 24, 2020
number theorydividesdivisibleGCD

Problem Statement

Let a,b,c,da, b,c,d be positive integers such that ac+bdac+bd is divisible by a2+b2a^2 +b^2. Prove that gcd(c2+d2,a2+b2)>1gcd(c^2 + d^2, a^2 + b^2) > 1.
Trần Nam Dũng
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