Symmetric Equations
Find four distinct positive integers such that:
and
Submit your answer as the sum of the squares of .
The answer is 39.
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If a>2 and b>2 and c>2 and d>2, then a + b < ab = c + d < cd, which is a contradiction. Therefore, WLOG we can assume that d is either 1 or 2. d = 1:
ab = a + b + 1
ab - a - b + 1 = 2
(a - 1)(b - 1) = 2, a = 3 and b = 2.
d = 2:
ab = a/2 + b/2 + 2
ab - a/2 - b/2 + 1/4 = 9/4
(2a - 1)(2b - 1) = 9
This holds no new distinct solutions for a and b.