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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.

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Assume that $a - b - c + d = 1$. Then you'll get from your result that

$$a^2 + b^2 + c^2 + d^2 = (a+b+c+d)(a-b-c+d) = a+b+c+d \tag{1}\label{eq1}$$

Now, since $a,b,c,d\in\mathbb{Z^+}$, note that $a^2 \gt a$ if $a \gt 1$, and likewise for $b, c, d$. Thus, \eqref{eq1} can only be true is $a = b = c = d = 1$. In that case, $a^2 + b^2 + c^2 + d^2 = 4$ which is composite. Of course, if $a - b - c + d \neq 1$, then it's also composite.

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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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