Yes, a b c d!

Number Theory Level 2

Let $a$ , $b$ , $c$ , and $d$ be positive integers such that $ad=b^2+bc+c^2$ . Is $a^2+b^2+c^2+d^2$ always composite?

No Yes

1 solution

Chew-Seong Cheong
Feb 27, 2019

$\begin{aligned} A & = {\color{#3D99F6}a^2} + b^2 + c^2 + \color{#3D99F6} d^2 \\ & = {\color{#3D99F6}(a+d)^2 - 2ad} + b^2 + c^2 & \small \color{#3D99F6} \text{Given that }ad = b^2 + bc + c^2 \\ & = (a+d)^2 - 2\left(\color{#3D99F6}b^2 + bc + c^2\right) + b^2 + c^2 \\ & = (a+d)^2 - \left(b^2 + 2bc + c^2\right) \\ & = (a+d)^2 - (b+c)^2 \\ & = (a+b+c+d)(a-b-c+d) \end{aligned}$

As brought out by @Jordan Cahn (see comments below), $A$ is not composite if $a-b-c+d = 1$ . But it is impossible to have $a-b-c+d = 1$ also satisfying $ad = b^2+bc+c^2$ and $a, b, c, d > 0$ .

Therefore, $a^2+b^2+c^2+d^2$ is always composite .

It is possible that $a-b-c+d=1$ , in which case it doesn't follow immediately that $A$ is composite.

It turns out that it's not possible to have $a-b-c+d=1$ while also satisfying $ad=b^2+bc+c^2$ and $a,b,c,d$ positive, but I don't think that's at all obvious.

Jordan Cahn - 2 years, 3 months ago

You are good.

Chew-Seong Cheong - 2 years, 3 months ago

Yes, a b c d!

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