Prove that there exists a unique prime number of the form $p^2 − 1$ where $p\geq 2$ is an integer.
I have no idea how to approach the question. any hints will be greatly appreciated
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Correct me if i´m wrong, but $\forall p \geq 3$ prime, $p$ is odd right? So if a prime have the following form $p^2-1$, the only available option would be $p=2$.
Willian Nunes
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I think the only requirement is that $p \in \mathbb{Z}$ and $p\ge 2$ – meiji163 Nov 30 '15 at 00:02
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The $p$ in the question is not an odd prime, or odd composite, although you do need to show this. So it has to be an even integer. – Daniel Buck Jul 03 '18 at 21:30
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Another hint: Do you remember how to factor the expression $a^2-b^2$?
And if you do, how does this relate to the expression $p^2-1$?
Mankind
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This question is asked so frequently on this website that the answer is immediately and unthinkingly coughed up in its shortest form each time.
What I'd like to do is show you one way you might be able to come up with the answer yourself if you don't already know it.
Does $p$ itself have to be prime? I will assume not, so I prefer to use the letter N instead of the letter P. Alright, then, can $n^2 - 1$ be prime?
Suppose $n$ is coprime to $3$. Then $n^2 \equiv 1 \pmod 3$ and therefore $n^2 - 1$ is a multiple of $3$. In every case except $n = -2$ or $2$, we will see that $n^2 - 1$ is a nontrivial multiple of $3$ (since the squares of negative numbers are positive numbers, I won't say anything further about negative numbers in this answer).
Then for $n^2 - 1$ to be prime with $n > 2$, $n$ has to be a multiple of $3$. If $n$ is an odd multiple of $3$, then $n^2$ is odd also but $n^2 - 1$ is an even composite number, that is, a nontrivial multiple of $2$.
Now the only possibility left is that $n$ is a multiple of $6$. Checking a few small cases, we see a pattern emerge:
- $6^2 - 1 = 35 = 5 \times 7$
- $12^2 - 1 = 143 = 11 \times 13$
- $18^2 - 1 = 323 = 17 \times 19$
- $24^2 - 1 = 575 = 5^2 \times 23$ but wait! That's $23 \times 25$.
This suggests $(n - 1)(n + 1)$. Applying FOIL, we get $$(n - 1)(n + 1) = n^2 + n - n - 1 = n^2 - 1.$$
Mr. Brooks
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$n(x) = n^2-1$
$n(x) = n^2-1^2$
$n(x) = (n+1)(n-1)$
For it to be prime it has to be only be divisible by itself. What to do from here?
danny13
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Assuming you're the OP, it would be appropriate to add these thoughts to the question body. :) – Andrew D. Hwang Nov 30 '15 at 02:31
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Regarding your question, you've shown that $n^{2} - 1$ factors into a product of integers. But by hypothesis, $n^{2} - 1$ is prime; what does this tell you about the factors in your factorization? – Andrew D. Hwang Nov 30 '15 at 02:40
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i would post this in the question but that was a guest account so it wont let me access anymore. This is an account with passowrd – danny13 Nov 30 '15 at 02:49
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$n^2-1$ factoring into a set of integers means its not prime and therefore disproves. Am I correct? – danny13 Nov 30 '15 at 02:51
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That's not quite correct; even a prime factors into a product of positive integers, just in a very particular way. – Andrew D. Hwang Nov 30 '15 at 11:30
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For it to be prime it has to be divisible by itself. I did the first couple numbers assuming n >=2 and none of them resulted in a prime, except 2. So I am thinking that n=2. But that is not a proof. So right now i dont know how to approach proving it – danny13 Nov 30 '15 at 16:56
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As a general, good-natured suggestion, it will help you formulate ideas clearly to avoid the pronoun "it". That aside, if $p$ is prime, what are the positive integers that divide $p$? And if $p = n^{2} - 1 = (n +1)(n - 1)$, what must $n$ be equal to as a consequence? – Andrew D. Hwang Nov 30 '15 at 22:49
