If we choose n as a square say $m^2$ then
$n(n+1) = m^2(m^2+1) = m^4 + m^2$
now n(n+1) is reperesented as sum of 2 squares that is $(m^2)^2 + m^2$
if $m^2$ can be represented as sum of 2 squares that is $p^2 + q^2$ this is possible as as (m,p,q) form a pythagorean triple then we have
$n(n+1) = (p^2+q^2)(m^2+1) = (pm+ q)^2 + (p-qm)^2$ this is another way
if we chose $p=x^2-y^2, q = 2xy, m= x^2+y^2, n = (x^2+y^2)^2 $ then it satisfies the condition (p,q,m) form a pythagrean triple)
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