# 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.

## 2 Answers

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.

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.