Question 869821
If we define our integers like this n-2,n-1,n,n+1,n+2, then most terms will cancel on either side of the equation when we square them
(n+1)^2 + (n+2)^2 = n^2 + (n-1)^2 + (n-2)^2
If you perform the multiplication, and collect terms you will be left with:
n^2 - 12n = 0
n(n-12) = 0
This gives two solutions, n=0 and n=12
So the integers are -2,-1,0,1,2 and 10,11,12,13,14