Challenge of Squares

Let three consecutive even numbers be $2n, 2n+2, 2n+4$

$(2n)^2+(2n+2)^2=(2n+4)^2$

$\Leftrightarrow n^2-2n-3=0$

$\Leftrightarrow n=-1, n=3$

Two sets of integers satisfy this condition: $(-2,0,2)$ and $(6,8,10)$ , whose sum are 0 and 24, consecutively.

It should have been more clearly stated that the numbers are positive.

let $a$ be the smallest even number. to find the three distinct consecutive even numbers: $(a)^2 + (a+2)^2 = (a + 4)^2$ $a^2 + a^2 + 4a + 4 = a^2 + 8a + 16$ $2a^2 + 4a + 4 = a^2 + 8a + 16$ $a^2 - 4a -12 = 0$ $a_1 = 6 ; a_2 = 2$ If you use $2$ , then you will get $2, 0, 2$ which is not consecutive. If you use $6$ , then you will get $6, 8, 10 (6^2 + 8^2 = 36 + 64 = 100)$ and $6+8+10 = \boxed {24}$