Hi,
How to show that if ab=cd $\displaystyle \Longrightarrow a^2+b^2+c^2+d^2$ is not prime????
let $\displaystyle \gcd(a,c)=x, \ \gcd(b,d)=y.$ then $\displaystyle a=xa_0, \ c=xc_0, \ b=yb_0, \ d=yd_0,$ where $\displaystyle \gcd(a_0,c_0)=\gcd(b_0,d_0)=1.$ thus, since $\displaystyle ab=cd,$ we must have $\displaystyle a_0=d_0$ and $\displaystyle b_0=c_0.$
therefore $\displaystyle a^2+b^2+c^2+d^2=(x^2+y^2)(a_0^2+b_0^2). \ \Box$
Let $\displaystyle p=(a,c)$, then $\displaystyle a=pr,\;c=ps,\;(r,s) = 1$
Since $\displaystyle ab=cd,\; prb=psd$, so $\displaystyle rb=sd. \; s|b $
Assume $\displaystyle b=qs$, from $\displaystyle rb=sd$ we get $\displaystyle rqs=sd$ or $\displaystyle d=qr$
Now put $\displaystyle a=pr,\; b=qs,\; c=ps,\; d=qr$ into $\displaystyle a^2+b^2+c^2+d^2$ we get
$\displaystyle a^2+b^2+c^2+d^2=(pr)^2+(qs)^2+(ps)^2+(qr)^2$
$\displaystyle =p^2 r^2+p^2 s^2+q^2 r^2+q^2 s^2 =(p^2+q^2)(s^2+r^2)$
Where $\displaystyle p^2+q^2, s^2+r^2 >1$ and so $\displaystyle a^2+b^2+c^2+d^2$ is not prime
I am curious if this statement is correct ...
If a prime can be expressed as the sum of two squares $\displaystyle x^2 + y^2 $, then $\displaystyle ( |x| , |y|) $ is the unique pair .
If it is true , we can solve it in this way :
$\displaystyle A = a^2 + b^2 + c^2 + d^2 = (a-b)^2 + (c+d)^2 = (a+b)^2 + (c-d)^2 $
If $\displaystyle a+b = c+d $ , then $\displaystyle a^2 + b^2 = c^2 + d^2 $ so $\displaystyle A $ is an even number ( not 2 , not prime ) .
Therefore , A has at least two pairs and it is not prime .