Thread: 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.

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

Originally Posted by CaptainBlack

In either case ac+db is not prime.

How does that show that is not prime?

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

Look. Are some people blind or what? The question is asking to show that – NOT – is not prime.

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.

Originally Posted by JaneBennet

How does that show that is not prime?

It doesn't, because I misread the question

RonL

Originally Posted by JaneBennet

Look. Are some people blind or what? The question is asking to show that – NOT – is not prime.

Wow, you're mean.

Originally Posted by Mathstud28

Wow, you're mean.

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

-Dan

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

Originally Posted by Mathstud28

[snip]....a mean individual

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

(*sigh* a desperate attempt at a statistics joke ...... )

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

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

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.

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

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


You have



This reduces to a quadratic equation for a:
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:
=>2 CAN'T be a solution
=>ab+cd=even 2=>ab+cd is not a prime

Hope that helps