Consecutive odd numbers

Assume, $a<b<c<d$ and let $a=2n+1$ where $n$ is an integer. Since $a,b,c,d$ are consecutive odd integers, it follows that $b=2n+3, c=2n+5, d=2n+7$ .

Observe that $a^2+b^2+c^2+d^2=(2n+1)^2+(2n+3)^2+(2n+5)^2+(2n+7)^2=(2n)^2+(2n)^2+(2n)^2+(2n)^2+2(2n)(1)+2(2n) (3)+2(2n)5 + 2(2n)7+1^2+3^2+5^2+7^2.$ It is quite obvious that the terms but $1^2+3^2+5^2+7^2$ have a factor of $4$ , it suffices to check if $1^2+3^2+5^2+7^2$ has a factor of $4$ and indeed it does as $1^2+3^2+5^2+7^2=84$ .