Jumat, 27 Agustus 2010

Prime Number problem (1)


08.47 | ,

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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