If $a,b,c$ and $d$ non-zero natural number such that $ab=cd$

Question

Question :

If $a,b,c$ and $d$ non-zero natural number such

that $ab=cd$

Show that :

$a^2+b^2+c^2+d^2$ is not prime number

My try :

Call $m$ : $\gcd$ of $a,b$ then $m|_a$ and $m|_b$

Then $\exists \alpha,\beta$ such that :

$a=m.\alpha$ and $b=m.\beta$

So $b\alpha=d\beta$

But how I complete ??

Answer 1

the intermediate conclusion from what you wrote is $$ m^2 \alpha \beta = cd $$ which does not seem to help.

Start over with $$ g = \gcd(a,c), $$ $$ a = g \alpha \; , $$ $$ c = g \gamma \; . $$ First, $$ \gcd(\alpha, \gamma) = 1 $$ From $ab=cd$ we have $g \alpha b = g \gamma d,$ so $$ \alpha b = \gamma d. $$ Since $\alpha,\gamma$ are coprime, we must have $\alpha |d.$ Write $$ d = h \alpha. $$ Then $$ \alpha b = \gamma \alpha h \; , $$ $$ b = h \gamma $$ Together $$ a = g \alpha, \; \; b = h \gamma, \; \; c = g \gamma, \; \; d = h \alpha \; . $$ Then $$ a^2 + b^2 + c^2 + d^2 = a^2 + c^2 + d^2 + b^2 = g^2 (\alpha^2 + \gamma^2) + h^2 (\alpha^2 + \gamma^2) = (g^2 + h^2) (\alpha^2 + \gamma^2) $$ Then $$ a^2 + b^2 + c^2 + d^2 = (g^2 + h^2) (\alpha^2 + \gamma^2) $$

Answer 2

$\begin{array}\\ n &=a^2+b^2+c^2+d^2\\ &=a^2+b^2+2ab+c^2+d^2-2cd\\ &=(a+b)^2+(c-d)^2\\ &=a^2+b^2-2ab+c^2+d^2+2cd\\ &=(a-b)^2+(c+d)^2\\ \end{array} $

If $a+b=c+d$ and $a-b=c-d$ then $a=c$ and $b=d$ so $n=2(a^2+b^2) $ so $n$ is not prime.

If $a+b\ne c+d$ and $a-b\ne c-d$ then $n$ can be written as the sum of two squares in two different ways and so is not prime.

Answer 3

We have $d=\frac{ab}{c}.$

Thus, $$a^2+b^2+c^2+d^2=a^2+b^2+c^2+\frac{a^2b^2}{c^2}=\frac{(a^2+c^2)(b^2+c^2)}{c^2}.$$ Can you end it now?

Answer 4

Let $(a,d)=e,\dfrac aA=\dfrac dD=e\ge1$ so that $(A,D)=1$

Similarly let $(b,c)=f, \dfrac bB=\dfrac cC=f;(B,C)=1$

$ab=cd\implies AB=CD$

$B=\dfrac{CD}A\implies A$ divides $C$

$D=\dfrac{AB}C\implies C$ divides $A$

$\implies A=\pm C\implies B=\pm D$

$a^2+b^2+c^2+d^2=e^2C^2+e^2D^2+f^2D^2+f^2C^2=(C^2+D^2)(e^2+f^2)$

Both factors are clearly $>1$