1. ## Proving.

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

2. ## Re: Proving.

Originally Posted by RuyHayabusa
sqrt(p+q)-sqrt(q) and p,q are natural numbers.prove that p is prime.
What? p = 2, q = 3, $\displaystyle \sqrt{2 + 3} - \sqrt{3} = \sqrt{5} - \sqrt{3}$ is definitely not prime. There is something missing in your problem statement. Please rewrite this with all information.

-Dan

3. ## Re: Proving.

Originally Posted by topsquark
What? p = 2, q = 3, $\displaystyle \sqrt{2 + 3} - \sqrt{3} = \sqrt{5} - \sqrt{3}$ is definitely not prime. There is something missing in your problem statement. Please rewrite this with all information.

-Dan
I meant IF sqrt(p+q)-sqrt(q) is natural and p and q are natural then prove that p is prime

4. ## Re: 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: $\displaystyle q = \dfrac{(k^2-p)^2}{4k^2}$.

5. ## Re: 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

6. ## Re: 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.

7. ## Re: Proving.

Originally Posted by emakarov
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.
Since q is a natural number, you can show k divides p. Since k divides p, you can show q is a perfect square. It just takes a little work.

8. ## Re: Proving.

Originally Posted by SlipEternal
$\displaystyle \sqrt{p+q} - \sqrt{q} = k$ for some natural number k. So $\displaystyle p = k(\sqrt{p+q} + \sqrt{q})$.

Solve for q: $\displaystyle q = \dfrac{(k^2-p)^2}{4k^2}$.
All this tells me p is divisible by k. That does not prove p is prime, unless k = p. Am I right

9. ## Re: 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

10. ## Re: Proving.

Originally Posted by votan
All this tells me p is divisible by k. That does not prove p is prime, unless k = p. Am I right
p = 12, q = 4. The statement is false. Q.E.D. :-P

11. ## Re: Proving.

Originally Posted by SlipEternal
p = 12, q = 4. The statement is false. Q.E.D. :-P
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.

12. ## Re: Proving.

Originally Posted by votan
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.
Reread the first post. The original statement says, Prove p is prime. I just proved p does not have to be prime.

13. ## Re: Proving.

Originally Posted by votan
The original question is

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

is a natural number and p is prime.
"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.

14. ## Re: Proving.

Originally Posted by emakarov
"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.
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.

15. ## Re: Proving.

Originally Posted by votan
Is the OP question incomlete, or misstated, let's see.
That's why I wanted the OP to respond first. Though I doubt (s)he's coming back at this point.

-Dan