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

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?

 
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)
 
Re: solutions to 2p + 1 = n^2

twisted_logic89 said:
My question is.... where do I begin?
Click to expand...


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?

 
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?
 
Re: solutions to 2p + 1 = n^2

twisted_logic89 said:
in n is a natural number then the factors of 2p would have to be greater than zero, right?
Click to expand...

You knew that from the assumption on p. In fact, I'd venture to say 2p >= 4 :wink:
 
\(\displaystyle p = \frac{1}{2}\cdot (n+1)\cdot (n-1)\)

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.
 
Re: solutions to 2p + 1 = n^2

twisted_logic89 said:
[If] n is a natural number then the factors of 2p would have to be greater than zero, right?
Click to expand...


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".


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


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 :)

 
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?
 
twisted_logic89 said:
... are the solutions for this 1 and 2?
Click to expand...


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.

 
twisted_logic89 said:
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 If you look at the column that I labeled 2p in my table above, then you will see that this product is not always even.

so... one could factor out a 2 from the original equation, correct? This statement is not worded correctly, but you're on the right track when n is odd ...

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?
Click to expand...
 
aah I am lost again, just when i thought I had it.
can anyone point me in the right direction? I am so close I can taste it.... and smell it a bit too....
 
twisted_logic89 said:
... can anyone point me in the right direction? ...
Click to expand...


I'm not sure that I know the answer to this question.

We know that 2p is the product of n - 1 and n + 1, for natural numbers n.

These two factors will always be two consecutive odd integers OR two consecutive even integers, dependent upon a specific n.

What happens when you multiply two even numbers? What happens when you multiply two odd numbers?

 
YES! ... was this an enthusiastic "you're on the right track" yes or an exasperated why-are-you-stating-the-obvious yes?

even * even = even
odd * odd = odd
 
twisted_logic89 said:
YES! ... was this an enthusiastic "you're on the right track" yes or an exasperated why-are-you-stating-the-obvious yes?

even * even = even
odd * odd = odd
Click to expand...

Look at the equation

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

p has been factorized - is it possible with a prime number?
 
twisted_logic89 said:
... [is] this an enthusiastic "you're on the right track" yes or an exasperated why-are-you-stating-the-obvious yes?
Click to expand...


The former.

Since you now realize that the product of n - 1 and n + 1 is even for some values of n, can any of those values of n lead to a product which is twice some prime number?

 
prime numbers cant be factorized right?

can any values of n lead to a product which is twice some prime number? since twice some prime number will be even, then only odd n values would work right?
 
twisted_logic89 said:
... since twice some prime number will be even ...
Click to expand...


2p = EVEN

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

Therefore, (n - 1) * (n + 1) must also be EVEN.

This is enough information to eliminate half of the natural numbers from being part of any solution.

Please tell me which natural numbers cannot possibly be part of any solution.

 
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