<s>[INST] If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite 
If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite.

My attempt so far:
$$a^2+b^2+c^2+d^2$$
$$=a^2+d^2+2ad+b^2+c^2+2bc-2ad-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+bc+c^2)-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+2bc+c^2)$$
$$=(a+d)^2+(b+c)^2-2(b+c)^2$$
$$=(a+d)^2-(b+c)^2$$
$$=(a+b+c+d)(a-b-c+d)$$
Now I am trying to prove that $a-b-c+d\not=1$ so I tried to assume the contradiction but I am unable to finish. Any help would be appreciated.
 [/INST] Firstly, if at least one of $a,b,c,d$ is greater than one than
$$a^2+b^2+c^2+d^2\gt a+b+c+d$$
$$\therefore a-b-c+d\gt 1$$
If they are all equal to $1$, then $1^2+1^2+1^2+1^2=4$ which is composite.
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