2021 IMO

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 {\displaystyle {\sqrt {\quad }}}2n <= x <= 2 * {\displaystyle {\sqrt {\quad }}}n At this time n >= 100, so

10 * {\displaystyle {\sqrt {\quad }}}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
at least the proposition is true