Prime Number problem (1)

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

= a²bd+ab²c+acd²+bc²d+ abcd-abcd

= bd ( a²-ac+c²) + ac ( b²+bd+d²)

= bd (b²+bd+d²) + ac ( b²+bd+d²) remember : a²-ac+c² = b²+bd+d²

= (bd+ac) ( b²+bd+d²) .................(1)

Note: let p a prime number, and GCD (b,p)=1 then, if dp divisible by b so d also divisible by b.

proof :

GCD (b,p) = 1, so there are m,n є Z so that:

1 = mb+np

then if two sides we multiply by d, we get

d = mbd+npd

= mbd+nkb ( because pd divisible by b)

= (md+nk) b

so, d divisible by b.

by used this property we can prove that (ab+cd) isn’t prime number. suppose (ab+cd) is a prime. so, (ac+bd) and (ad+bc) relative prime to (ab+cd).

from (1) we know if (ab+cd)(ad+bc) divisible by (ac+bd). so (ac+bd) must devide (ad+bc). this is contradiction with (ac + bd) > (ad + bc) .

so, (ab+cd) isn’t prime number.

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