# Thread: find all solutions to 2p + 1 = n^2, for n natural, p prime

1. ## find all solutions to 2p + 1 = n^2, for n natural, p prime

if n is a natural number and p is a prime number find all solutions to 2p + 1 = n^2

2. ## Re: solutions to 2p + 1 = n^2

Did I already give you my boilerplate response for posters who don't show any work, don't explain what they already know, and don't ask any questions?

3. ## Re: solutions to 2p + 1 = n^2

no, but I figured I would get an answer like that, so you really don't have to.

My question is.... where do I begin? My professor said he had not taught us what we needed to know to fully complete the problem (and therefore made it extra credit)

would the division algorithm be used? (a= qb + r)

4. ## Re: solutions to 2p + 1 = n^2

Originally Posted by twisted_logic89

My question is.... where do I begin?

2p = n[sup:391bq5es]2[/sup:391bq5es] – 1

2p = (n + 1) * (n – 1)

If n is a natural number, what can we say about the factors of 2p above?

5. ## Re: solutions to 2p + 1 = n^2

in n is a natural number then the factors of 2p would have to be greater than zero, right?

6. ## Re: solutions to 2p + 1 = n^2

Originally Posted by twisted_logic89
in n is a natural number then the factors of 2p would have to be greater than zero, right?
You knew that from the assumption on p. In fact, I'd venture to say 2p >= 4

7. ## Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

$p = \frac{1}{2}\cdot (n+1)\cdot (n-1)$

Now my question is -

What is the definition of a prime number?

Then my request is - after you found the definition - sit and think about the step above.

8. ## Re: solutions to 2p + 1 = n^2

Originally Posted by twisted_logic89

[If] n is a natural number then the factors of 2p would have to be greater than zero, right?

If you think of the expression 2p as twice some prime number, then my answer to your question is clearly "yes".

If you think of the expression 2p as n[sup:1278fexe]2[/sup:1278fexe] - 1, for any natural number n, then my answer to your question is "not always".

$\begin{tabular}{|c|c|c|c|} \hline n & n - 1 & n + 1 & 2p & \hline 1 & 0 & 2 & 0 & 2 & 1 & 3 & 3 & 3 & 2 & 4 & 8 & 4 & 3 & 5 & 15 & 5 & 4 & 6 & 24 & : & : & : & : & \hline \end{tabular}$

Subhotosh gave you the important question to consider; here are some more questions to consider.

How many prime numbers are even?

Is the value of 2p even or odd when p is any real number?

Is the product (n - 1) * (n + 1) even or odd when n is a natural number?

When n is odd, can you factor out a number from the product (n - 1) * (n + 1)?

If so, then can you simplify the original equation when n is odd?

~ Mark

9. ## Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

prime number- divided evenly only by itself and 1

the only even prime number is 2

2p will be even

product of (n-1)*(n+1) is even

so... one could factor out a 2 from the original equation, correct?

p= (1/2) * (n+1) * (n-1)

! i feel like I am getting really close with this...
are the solutions for this 1 and 2?

10. ## Re: find all solutions to 2p + 1 = n^2, for n natural, p prime

Originally Posted by twisted_logic89

... are the solutions for this 1 and 2?

Hmmm, it seems to me that stating the numbers 1 and 2 as solutions for an equation that contains two different symbols is somewhat ambiguous.

If you intend these two numbers to be p and n, then I suppose that you mean n = 1 and p = 2.

n = 1

n[sup:2rg8x8lu]2[/sup:2rg8x8lu] = 1

Is there a prime number p such that 2p + 1 = 1?

I don't think so. (It's certainly not p = 2.)

Maybe you meant n = 1 and n = 2 as solutions.

n = 2

n[sup:2rg8x8lu]2[/sup:2rg8x8lu] = 4

Is there a prime number p such that 2p + 1 = 4?

I don't think so.

I don't think 1 and 2 are solutions.

Keep trying.

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