# Prove that ab + cd is not prime.

• Apr 24th 2008, 08:05 AM
perash
Prove 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, 02:19 PM
CaptainBlack
Quote:

Originally Posted by perash
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.

b+d+a-c>0, so either ac+bd=0 and b+d-a+c=0, or ac+db>0 and b+d-a+c>0

So we have either ac+bd=0, or there exist positive integers u, v such that ac+db=uv. In either case ac+db is not prime.

RonL
• Jun 13th 2008, 07:06 PM
JaneBennet
Quote:

Originally Posted by CaptainBlack
In either case ac+db is not prime.

How does that show that $\color{red}ab+cd$ is not prime?
• Jun 14th 2008, 07:34 AM
abes22
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, 08:05 AM
JaneBennet
Look. Are some people blind or what? The question is asking to show that $ab+cd$NOT $ac+bd$ – is not prime.

Quote:

Originally Posted by perash
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.

• Jun 14th 2008, 10:44 AM
CaptainBlack
Quote:

Originally Posted by JaneBennet
How does that show that $\color{red}ab+cd$ is not prime?

It doesn't, because I misread the question

RonL
• Jun 14th 2008, 12:03 PM
Mathstud28
Quote:

Originally Posted by JaneBennet
Look. Are some people blind or what? The question is asking to show that $ab+cd$NOT $ac+bd$ – is not prime.

Wow, you're mean. (Rofl)
• Jun 14th 2008, 02:12 PM
topsquark
Quote:

Originally Posted by Mathstud28
Wow, you're mean. (Rofl)

She's not mean she's speaking her mind. I have a great deal of respect for those who do so.

-Dan
• Jun 14th 2008, 04:50 PM
Mathstud28
Quote:

Originally Posted by topsquark
She's not mean she's speaking her mind. I have a great deal of respect for those who do so.

-Dan

You are right, she is just an individual speaking her mind!....a mean individual :p
• Jun 14th 2008, 05:02 PM
mr fantastic
Quote:

Originally Posted by Mathstud28
[snip]....a mean individual :p

I think JB is a bit more than average ......

(*sigh* a desperate attempt at a statistics joke ...... )
• Jun 16th 2008, 02:10 AM
abes22
mrs 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, 03:03 AM
CaptainBlack
Quote:

Originally Posted by abes22
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

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, 03:46 AM
abes22
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, 05:33 AM
topsquark

-Dan
• Jun 30th 2008, 08:15 AM
fobos3
$a x^2+b x+c=0$
$x=\frac{-b\pm \sqrt{b^2-4 a c}}{2 a}$

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

You have
$a c+b d=(a+b-c+d) (-a+b+c+d)$
$a c+b d=a^2-b^2+c^2-d^2$

This reduces to a quadratic equation for a: $a^2-\text{ac}-b^2+c^2+b d-d^2=0$
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:
$\text{ab}+\text{cd}=4*3+2*1\neq 2$=>2 CAN'T be a solution
=>ab+cd=even $\neq$2=>ab+cd is not a prime

Hope that helps