Prime 2

**bhitroofen01** · May 1st 2010, 06:11 AM

Hi,

How to show that if ab=cd $\Longrightarrow a^2+b^2+c^2+d^2$ is not prime????

**elim** · May 1st 2010, 02:57 PM

Are you sure this is true?

**Deadstar** · May 1st 2010, 03:13 PM

Originally Posted by elim

Are you sure this is true?

Do you know a counterexample? I can't think of one myself.

**NonCommAlg** · May 1st 2010, 07:51 PM

Originally Posted by bhitroofen01

Hi,

How to show that if ab=cd $\Longrightarrow a^2+b^2+c^2+d^2$ is not prime????

let $\gcd(a,c)=x, \ \gcd(b,d)=y.$ then $a=xa_0, \ c=xc_0, \ b=yb_0, \ d=yd_0,$ where $\gcd(a_0,c_0)=\gcd(b_0,d_0)=1.$ thus, since $ab=cd,$ we must have $a_0=d_0$ and $b_0=c_0.$

therefore $a^2+b^2+c^2+d^2=(x^2+y^2)(a_0^2+b_0^2). \ \Box$

**elim** · May 1st 2010, 08:02 PM

Let $p=(a,c)$ , then $a=pr,\;c=ps,\;(r,s) = 1$

Since $ab=cd,\; prb=psd$ , so $rb=sd. \; s|b$

Assume $b=qs$ , from $rb=sd$ we get $rqs=sd$ or $d=qr$

Now put $a=pr,\; b=qs,\; c=ps,\; d=qr$ into $a^2+b^2+c^2+d^2$ we get

$a^2+b^2+c^2+d^2=(pr)^2+(qs)^2+(ps)^2+(qr)^2$

$=p^2 r^2+p^2 s^2+q^2 r^2+q^2 s^2 =(p^2+q^2)(s^2+r^2)$

Where $p^2+q^2, s^2+r^2 >1$ and so $a^2+b^2+c^2+d^2$ is not prime

**simplependulum** · May 1st 2010, 08:34 PM

I am curious if this statement is correct ...

If a prime can be expressed as the sum of two squares $x^2 + y^2$ , then $( |x| , |y|)$ is the unique pair .

If it is true , we can solve it in this way :

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

If $a+b = c+d$ , then $a^2 + b^2 = c^2 + d^2$ so $A$ is an even number ( not 2 , not prime ) .

Therefore , A has at least two pairs and it is not prime .

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