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A problem I saw on quora, slightly modified.

If $a, b, c, d \in \mathbb{N}$ and $ab = cd \ne 0$ show that $a^2+b^2+c^2+d^2$ is composite.

My solution uses the theorem that a number that can be written as the sum of two squares in two different ways is composite, and I wondered if there is a simpler proof (probably involving the difference of squares).

Here is my proof.

$a^2+b^2+c^2+d^2 =a^2+2ab+b^2+c^2–2cd+d^2 =(a+b)^2+(c-d)^2 $

and

$a^2+b^2+c^2+d^2 =a^2-2ab+b^2+c^2+2cd+d^2 =(a-b)^2+(c+d)^2 $

and it is known that a number that can be written as the sum of two squares in two different ways is composite.

If the two ways are the same then either $a+b=a-b, c-d=c+d$ or $a+b=c+d, c-d=a-b$.

In the first case, $b=d=0$ which is not allowed.

In the second case, adding them $2a = 2c$ so $a=c$ and then $b=d$ so $a^2+b^2+c^2+d^2 =2a^2+2b^2 $ which is composite.

Bill Dubuque
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marty cohen
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1 Answers1

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standard is to let $g = \gcd(a,c)$ so that $a = g \alpha$ and $c = g \gamma,$ with $$ \gcd(\alpha, \gamma) = 1 $$

As $b \alpha = d \gamma$ and $ \gcd(\alpha, \gamma) = 1, $ we find $\gamma|b,$ let $ b = h \gamma,$ which leads to $d = h \alpha$

$$ a^2 + c^2 + d^2 + b^2 = g^2 \alpha^2 + g^2 \gamma^2 + h^2 \alpha^2 + h^2 \gamma^2 = (g^2 + h^2) (\alpha^2 + \gamma^2) $$

Will Jagy
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