0

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 ??

kccu
  • 20,808
  • 1
  • 22
  • 41
Kînan Jœd
  • 731
  • 1
    By "neutral" do you mean "natural"? – Robert Israel May 16 '19 at 03:03
  • Yes , sorry sir , thanks – Kînan Jœd May 16 '19 at 03:06
  • Note that we write $m \mid a$ for "$m$ divides $a$", not $m|_a$. Also you should use "\cdot" ($\cdot$) for multiplication, not a period. In addition, I don't see how you go from $ab=cd$ and $a=m\alpha$, $b=m\beta$ to $b\alpha=d\beta$? – kccu May 16 '19 at 03:25
  • 1
    Hint: $a^2+b^2+c^2+d^2 = (a+b)^2+(c-d)^2$ – Rushabh Mehta May 16 '19 at 03:29
  • @Don Thousand: I thought of that also, and used it as the basis for my proof, so I upvoted your comment. – marty cohen May 16 '19 at 03:37
  • $ab=cd$ implies that the vectors $\vec{u}=(a,c)$ and $\vec{v}=(b,d)$ are parallel (=linearly dependent over $\Bbb{Q}$, hence over $\Bbb{Z}$). Therefore they are both integer multiples of $\vec{w}=\dfrac{\gcd(a,b)}a\vec{u}$. The claim follows from this as $||\vec{w}||^2$ is a factor of both $||\vec{u}||^2$ and $||\vec{v}||^2$. – Jyrki Lahtonen May 16 '19 at 04:43

4 Answers4

4

$\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.

marty cohen
  • 107,799
1

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?

Michael Rozenberg
  • 194,933
1

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 $>1$

lab bhattacharjee
  • 274,582
0

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

Will Jagy
  • 139,541
  • I'd like to add that this is an equivalence: $ab=cd$ iff there exist $x,y,z,t$ such that $a=xy$, $b=zt$, $c=xt$, $d=yz$. – sdcvvc May 17 '19 at 09:38