sqrt(p+q)-sqrt(q) and p,q are natural numbers.prove that p is prime.

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- Oct 9th 2013, 06:53 AMRuyHayabusaProving.
sqrt(p+q)-sqrt(q) and p,q are natural numbers.prove that p is prime.

- Oct 9th 2013, 07:10 AMtopsquarkRe: Proving.
- Oct 9th 2013, 07:19 AMRuyHayabusaRe: Proving.
- Oct 9th 2013, 08:24 AMSlipEternalRe: Proving.
$\displaystyle \sqrt{p+q} - \sqrt{q} = k$ for some natural number k. So $\displaystyle p = (\sqrt{p+q} -\sqrt{q})(\sqrt{p+q}+\sqrt{q}) = k(\sqrt{p+q} + \sqrt{q})$.

Solve for q: $\displaystyle q = \dfrac{(k^2-p)^2}{4k^2}$. - Oct 9th 2013, 08:45 AMtopsquarkRe: Proving.
Just to prove I'm a jerk... I still think the problem statement is awfully fuzzy. I'm sure SlipEternal got it, so no worries, but there should be something in the problem statement that says "show that you can pick a value for q" etc.

Ah well.

-Dan - Oct 9th 2013, 09:25 AMemakarovRe: Proving.
The confusion in the problem statement arises from the strange enumeration: "... and p, q are ...". It would be clearer if the problem said, "..., p and q are ..." or "... as well as p and q are ...". In any case, the problem says:

Given: $\displaystyle \sqrt{p+q}-\sqrt{q}$, p and q are natural numbers. Prove: p is prime.

I am not sure how post #4 helps to solve it. - Oct 9th 2013, 11:26 AMSlipEternalRe: Proving.
- Oct 9th 2013, 11:45 AMvotanRe: Proving.
- Oct 9th 2013, 11:54 AMtopsquarkRe: Proving.
I'd like to stop for a moment and see what RuyHayabusa has to say about the problem statement considering the confusion of the previous posts. This is mostly speculation about what (s)he meant.

-Dan - Oct 9th 2013, 01:17 PMSlipEternalRe: Proving.
- Oct 9th 2013, 03:52 PMvotanRe: Proving.
I am still confused. First you used the wrong example choosing p=12. p must be prime. The proof by negation as provided is missing something.

The original question is

A = sqrt(p+q) - sqrt(q)

is a natural number and p is prime. For this to be true, p+q and q both must be perfect squares.

Take these two pairs of numbers (p,q): (11,25) and (13,36)

In both cases A and B are prime numbers; B real conjugate of A. If k is that some natural number proposed by SlipEternal, then we will have

p = kB ==> p/k = B

but p is prime how you could divide it by any other number other than by itself. Even for k=1, p is not B and the two test pairs of numbers used here prove that.

I am not a number theorist, but I don't see how the provided proof is a proof for some natural k when by definition p cannot be divided by other than p. - Oct 9th 2013, 04:00 PMSlipEternalRe: Proving.
- Oct 9th 2013, 04:13 PMemakarovRe: Proving.
"That was not a question, it was a comment" (C) Dirac :)

The original problem, as I understood it, was to*prove*that p is prime (thus, post #10 shows that the statement does not hold), but the OP can clarify it if he/she wants. - Oct 9th 2013, 04:57 PMvotanRe: Proving.
The OP restated the question in post 3: "I meant IF sqrt(p+q)-sqrt(q) is natural and p and q are natural then prove that p is prime"

This is my contention, the question is to prove, and the proof was to deny. Is there a way of proving the OP question, I couldn't get there. Is the OP question incomlete, or misstated, let's see. I don't see what is the point of proving p not prime. To me it is obvious. - Oct 10th 2013, 06:29 AMtopsquarkRe: Proving.