2021/091) if ab = cd prove that $a^2+b^2 + c^2 + d^2 $ is composite

 We have

ab = cd

or $\frac{a}{c}= \frac{d}{b} = \frac{m}{n}$ where m and n are in lowest terms or gcd(m,n) = 1

so an = cm and dn = bm

now $n^2(a^2+b^2+ c^2+d^2) $

$= (na)^2 + (nb)^2 + (nc)^2 + (nd)^2$

$= (cm)^2 + (bn)^2 +(nc)^2 + (nd)^2= (b^2+c^2)(m^2 + n^2)$

or $a^2+b^2+c^2+d^2 = \frac{b^2+c^2}{n^2} (m^2+n^2)$

as n is less than $b^2+c^2$ so it is product of 2 numbers > 1 so composite