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Suppose that $$r=\frac{p^m+1}{2}$$ is a prime number, where $p$ is also prime. Does the equation $$p^{2}-2\equiv 0 \pmod{r}$$ have any solutions?

Thanks in advance.

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    Have a look at this http://en.wikipedia.org/wiki/Legendre_symbol and this http://en.wikipedia.org/wiki/Law_of_quadratic_reciprocity2012-06-05
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    Is the $p$ in the definition of $r$ the same as the $p$ in the equation? In other words, are you asking whether there are any primes $p$ such that **both** $(p^m+1)/2$ is a prime, *and* $p^2\equiv 2\pmod{(p^m+1)/2}$ holds? Or are you just asking whether $2$ is a square modulo $(p^m+1)/2$ when both $p$ and $(p^m+1)/2$ are primes?2012-06-05
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    @Arturo I think the latter $p$ is meant to be $x$, given the title of the post.2012-06-05
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    @alex.jordan: That was not the original title. If you look at the edit history, you will see that the original title was **Question in number theory.** That title was put in by M Turgeon in his edit.2012-06-06
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    Are there any conditions on $m$? Or certainly there will be counterexamples.2012-06-07
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    @awllower No, but it is clear that $m$ is even.2012-06-09
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    Two solutions were posted several days ago, Sara. Do you have nothing to say about either one?2012-06-10
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    @Gerry Myerson: Thank you very much for your interest and answer. Now everything is fixed with my question, and thank you once again.2012-06-10
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    I don't know what you mean by "everything is fixed with my question". If you mean that one of the answers given here has answered your question, you should consider accepting that answer. If you mean you found an answer without help from the answers here, you should consider letting people here know what was wrong with their answers. And if you mean something else, what?2012-06-11
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    @Gerry Myerson: I mean that, my problem is solved by your answer. Your answer was the best for me.2012-06-11
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    Good, glad to know that. The standard way to indicate that on this website is to "accept" the answer by clicking in the little check-mark next to it.2012-06-11

3 Answers 3

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If you are asking whether $2$ is a quadratic residue of $r$, it is easy to find by treating each of the following 8 cases: when $p \equiv 1, 3, 5, 7, 9, 11, 13, 15 (\bmod\, 16)$ respectively. Suppose $r=(p^m+1)/2$ is a prime, then the condition for $2$ being a quadratic residue of $r$ is

when $p=16n+1$, $m$ can be any natural number

when $p=16n+3$ or $16n+11$, $m\equiv 0\, (\bmod\, 4)$

when $p=16n+5$, $m\equiv 0\ \mbox{or}\ 3\, (\bmod\, 4)$

when $p=16n+7$ or $16n+9$ or $16n+15$, $m$ can be any even number

when $p=16n+13$, $m\equiv 0\ \mbox{or}\ 1\, (\bmod\, 4)$

The principle is to raise the least residues to (at most) power $4$, plus $1$ then divide by $2$, and find those which $\equiv \pm 1\, (\bmod\, 8)$

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    If $p=16n+1$. We can $p=17$ for example, then $(p^m +1)/2$ when $m$ is odd number is not prime. In my question $(p^m +1)/2$ is prime.2012-06-10
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    You are right. But I have already supposed that $(p^m+1)/2$ is a prime number.2012-06-11
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For any prime $\,q\,$, the equation $\,x^2=2\pmod q\,$ has a solution iff $\,q=\pm 1\mod 8\,$ . As I am assuming that in "$\,p^2-2=0\pmod r\,$" you actually meant $\,x^2-2=0\pmod r\,$ (if I'm wrong in my assumption please disregard this post), I'll give you two examples with different outcomes:

== for $\,\displaystyle{\,p=3\,,\,m=4\,,\,r=\frac{3^4+1}{2}=41=1\pmod 8\,\,}$ , so your equation has solution

== for $\,\,\displaystyle{p=5\,,\,m=2\,\,,\,r=\frac{5^2+1}{2}=13\neq \pm1\pmod 8}\,\,$ , and your equation has no solution here.

Thus I believe you must either change your question or impose further conditions.

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I'm going to assume the question as posed is the question intended.

$p^2-2\equiv0\pmod r$ requires $r\lt p^2$. But $r=(p^m+1)/2$ almost requires $r\gt p^2$, so the two relations are nearly mutually exclusive. The only ways they can both hold are

  1. $m=1$. Then $r=(p+1)/2$; $p=2r-1$; $p^2-2=4r^2-4r-1\equiv-1\pmod r$, so no solutions here.

  2. $m=2$. Then $r=(p^2+1)/2$; $p^2=2r-1$; $p^2-2=2r-3\equiv-3\pmod r$; so the only way this is going to work is if $-3\equiv0\pmod r$, which says $r=3$, but this leads to $p=\pm\sqrt5$, which is nonsense, so no solutions here, either.

And that's it. If $m\ge3$ then you'll find that $r=(p^m+1)/2\gt p^2$, so there are no other cases to look at.

In summary, with the problem as posed, the congruence has no solutions.