# 2021 IMO

Revision as of 11:02, 29 January 2023 by Mathhyhyhy (talk | contribs)

For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that p+q = x^2 and q+r = y^2, p+r = z^2.

WLOG n<= p<= q<= r <= 2n ... Equation 1

p = x^2 + z^2 - y^2

q = x^2 + y^2 – z^2

r = y^2 + z^2 – x^2

by equation 1

2n <= x^2 + z^2 – y^2 <= 4n

2n <= x^2 + y^2 – z^2 <= 4n

2n <= y^2 + z^2 – z^2 <= 4n

6n <= x^2 + y^2 + z^2 <= 12n

6n <= 3x^2 <= 12n

2n <= x^2 <= 4n

√(2n) <= x <= 2√n

At this time n >= 100, so

10 * √2 <= x,y,z <= 20

15 <= x,y,z <= 20

where

2|x^2 + y^2 – z^2 2|x^2 + z^2 – y^2 2|y^2 + z^2 – z^2 x = 16, y = 18, z = 20 fits perfectly

therefore the proposition is true