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Algebra Difficulty 4.4 AIME Prove it Ireland

Show that from eight consecutive integers one can always pick four whose squares have the same sum as the squares of the remaining four.

Solution

Let the given eight consecutive integers be $n, n+1, \dots, n+7$ . Experimenting with the case $n=0$ leads to the identity
$n^2 + (n+3)^2 + (n+5)^2 + (n+6)^2 = (n+1)^2 + (n+2)^2 + (n+4)^2 + (n+7)^2.$
Alternatively, one may observe that
$n^2 + (n+3)^2 - (n+1)^2 - (n+2)^2 = 4.$
Subtracting from it the same identity with $n$ replaced by $n+4$ yields the above identity.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.