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

[mmm4444bot]

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?

[twisted_logic89]

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]

2p is even when n is odd and odd when n is even...right?

[mmm4444bot]

... can anyone point me in the right direction? ...

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?

[mmm4444bot]

2p is even when n is odd and odd when n is even...right?

YES!

[twisted_logic89]

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

[Subhotosh Khan]

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

Look at the equation

[tex]p \, = \, \frac{1}{2}\cdot (n+1)\cdot (n-1)[/tex]

p has been factorized - is it possible with a prime number?

[mmm4444bot]

... [is] this an enthusiastic "you're on the right track" yes or an exasperated why-are-you-stating-the-obvious yes?

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?

[twisted_logic89]

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?

[mmm4444bot]

... since twice some prime number will be even ...

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.