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Saturday, November 6, 2021

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 

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