Maths Olympiad Prep

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Number theory Difficulty 4.8 AIME Prove it New Zealand

Problem:
Let aa and bb be positive integers with no common factor greater than 11. What are the possible values for the greatest common divisor of (a+b)(a + b) and (ab)(a - b)?

Solution

Solution:
Let dd be a common divisor of both (a+b)(a + b) and (ab)(a - b), and therefore divides linear combinations of (a+b)(a + b) and (ab)(a - b). In particular,
d[(a+b)+(ab)]=2a,d[(a+b)(ab)]=2b. d \mid [(a + b) + (a - b)] = 2a,\quad d \mid [(a + b) - (a - b)] = 2b.
As aa and bb don't share any common factors, the only common factors 2a2a and 2b2b have is 22. Hence d2d \mid 2, which means that d=1d = 1 or 22. This means that the greatest common divisor cannot be larger than 22.

We now show that both 11 and 22 can both occur.

Consider a=2a = 2, b=1b = 1
gcd(2+1,21)=1. \gcd (2 + 1, 2 - 1) = 1.
Consider a=5a = 5, b=3b = 3
gcd(5+3,53)=2. \gcd (5 + 3, 5 - 3) = 2.
Thus the only possible values for the greatest common divisor of (a+b)(a + b) and (ab)(a - b) are 11 and 22.

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