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Some problem of number theory

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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.