XXIX Rioplatense Mathematical Olympiad

Let us compute the sum $S$ of the squares of three consecutive odd numbers $x$, $x+2$, $x+4$: $$S=x^2+(x+2{)}^2+(x+4{)}^2=3x^2+12x+20.$$ We can see that $S$ is odd: indeed, $x$ is odd, hence $3x^2$ is odd, and $12x$ and $20$ are both even. Furthermore, $S$ has a remainder of 2 upon division by 3 (because $S=3(x^2+4x+6)+2$). Among the numbers 1111, 2222, 3333, ..., 9999, the only one that satisfies both conditions is 5555. Hence $3x^2+12x+20=5555$, and solving for $x$ yields $x=41$. Therefore, the only solution is $41^2+43^2+45^2=5555$.