2017/028) Find 3 consecutive numbers that can be expressed as sum of 2 non-zero squares

We have $(2n)^2+ (2n)^2 = 8n^2$
and $(2n+1)^2 + (2n-1)^2 = 8n^2 + 2$
If we can find $8n^2 + 1  = a^2+ b^2$ for some a, b  then we are through.
we haave
$(2n - a)^2   +   (2n + a - 1)^2   =   8n^2   +   1   -   (4n - (2a^2 - 2a))$
If we can have  n   and   a   such that
$4n   -   (2a^2 - 2a)   =   0$ then we are through
or $2n   =   a^2 - a   =   a(a - 1)$
or $n = \frac{a(a-1)}{2}$

as there are infinite number of a's we find infinite nuber of solutions

a = 2, gives n= 1 2n-a = 0 so we put
a=3 , gives n= 3 we get

$6^2+6^2 = 72$
$3^2 + 8^2 = 73$
$5^2+7^2 = 74$
similarly we can find more solutions by putting any value of a.