Let
a > b > c > d are natural numbers and ac + bd = (b + d + a –
c) (b + d – a +c). prove that ab + cd isn’t prime number !
Solutuon :
to prove this problem we wiil use contradiction.
we have :
a > b its meaning (a-b) > 0
c > d its meaning (c-d) > 0
so, (a-b) (c-d) = (ac-ad-bc+bd) > 0 so that (ac+bd) > (ad + bc) ...........(2)
similarly, for a>d and b> c we get (ab+cd) > (ac + bd) .........(3)
from (2) and (3) we get : (ab+cd) > (ac + bd) > (ad + bc)
from problem :
ac + bd = (b + d + a – c) (b + d – a +c)
= b2+bd–ab+bc+bd+d2-ad+cd +ab+ad-a2+ac-bc-cd+ac-c2
= b2+2bd+d2-a2+2ac-c2
a2-ac+c2 = b2+bd+d2
see that :
(ab+cd) (ad+bc) = a2bd+ab2c+acd2+bc2d
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