Let a, b, c, d

be integers with a > b > c > d > 0.

Suppose that

ac + bd = (b + d + a − c)(b + d − a + c).

Prove that ab + cd is not prime.

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- Apr 24th 2008, 07:05 AMperashProve that ab + cd is not prime.
Let a, b, c, d

be integers with a > b > c > d > 0.

Suppose that

ac + bd = (b + d + a − c)(b + d − a + c).

Prove that ab + cd is not prime. - Apr 24th 2008, 01:19 PMCaptainBlack
- Jun 13th 2008, 06:06 PMJaneBennet
- Jun 14th 2008, 06:34 AMabes22answer
if ac + bd = (b+d+a-c)(b+d-a+c)

we can factorise it, because (b+d+a-c)(b+d-a+c) is a difference of perfect squares

(b+d)^2-(a-c)^2

expands to b^2+d^2-a^2-c^2+2bd+2ac

factor of 2 here

b^2+d^2-a^2-c^2+2(a+b+c+d)

because there is a factor of two, the number is even and therefore cannot be prime - Jun 14th 2008, 07:05 AMJaneBennet
- Jun 14th 2008, 09:44 AMCaptainBlack
- Jun 14th 2008, 11:03 AMMathstud28
- Jun 14th 2008, 01:12 PMtopsquark
- Jun 14th 2008, 03:50 PMMathstud28
- Jun 14th 2008, 04:02 PMmr fantastic
- Jun 16th 2008, 01:10 AMabes22mrs mean
well if speaking ur mind is so acceptable, u r mean. in fact ur terrible. the question is to prove it true, but it does prove it. there is a factor of two in the entire thing. have a god day

- Jun 16th 2008, 02:03 AMCaptainBlack
The question is given

a, b, c, d integers with a > b > c > d > 0, and:

ac + bd = (b + d + a − c)(b + d − a + c).

Prove that ab + cd is not prime.

What you have proven is that ac + bd is not prime not that ab + cd is not prime. Which was the whole point of Jane's first post.

It is an easy mistake to make, I know because I made it myself.

(Also your argument could do with some work it is not entirly clear what you are trying to do. Also maybe your language could do with some moderation as well it is close to getting an infraction for being insulting)

RonL - Jun 16th 2008, 02:46 AMabes22
you're right, although then again I'm 15.

But that lady whoever she may be had no right to say something like that. Regardless of whether or not i did misread the question. So my insult was equally as bad, and I wasn't prepared to let her go unnoticed. - Jun 16th 2008, 04:33 AMtopsquark
I think enough has been said about this. Please only add a post here if you can contribute to the problem that has been asked.

-Dan - Jun 30th 2008, 07:15 AMfobos3
$\displaystyle a x^2+b x+c=0$

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4 a c}}{2 a}$

Therefore if (a,b,c,x) are integers, than x is divided by 2. $\displaystyle [1]$

You have

$\displaystyle a c+b d=(a+b-c+d) (-a+b+c+d)$

$\displaystyle a c+b d=a^2-b^2+c^2-d^2$

This reduces to a quadratic equation for a:$\displaystyle a^2-\text{ac}-b^2+c^2+b d-d^2=0$

So a is an even number (from 1). The same can be proved for any of the (a,b,c,d).

Because of that a*b+c*d=even*even+even*even=even+even=even

Even numbers are not primes except for 2. But the smallest values of a,b,c,d are:

d=1

c=2

b=3

a=4

So:

$\displaystyle \text{ab}+\text{cd}=4*3+2*1\neq 2$=>2 CAN'T be a solution

=>ab+cd=even$\displaystyle \neq$2=>ab+cd is not a prime

Hope that helps