if one of $a,b,c,d$ is equeal to $0$ we are done. Let $a\geqslant b\geqslant c\geqslant d>0$ If $b^2+c^2+d^2>N$ by vieta jumping we can assume that $a\leqslant \frac{Nbcd}{2}$ but then: $Nabcd+N=a^2+b^2+c^2+d^2\leqslant \frac{Nabcd}{2} +b^2+c^2+d^2\Rightarrow a\leqslant \frac{2(b^2+c^2+d^2-N)}{Nbcd}< \frac{6b^2}{Nbcd}=\frac{6}{Ncd}b\leq \frac{6}{Ncd}a\Rightarrow Ncd<6$ So $N<6$ with means that it can be writen as a sum of tree squares. If $N=b^2+c^2+d^2$ we are done. If $N>b^2+c^2+d^2$ $a^2+b^2+c^2+d^2=Nabcd+N>Nabcd+b^2+c^2+d^2\Rightarrow a>Nbcd$ Let $a=Nbcd+x$ then we have: $(Nbcd+x)^2+b^2+c^2+d^2=(Nbcd+x)Nbcd+N\Rightarrow xNbcd+x^2+b^2+c^2+d^2=N$ contradiction