For which primes p is 3 a quadratic residue modulo p?  From Quadratic Reciprocity -1/p = 1 if p = 1 mod 4 means that -1 is a quadratic residue mod p iff p = 1 mod 4. Similarly I know that I'm supposed to use 3/p = (p/3)(-1)^((p-1)/2) but how do I convert this to a similar form that gives the desired value of p?
40670
$(\frac 3p)$ is the Legendre symbol here, not a fraction, if that wasn't clear.  This is what I have so far:  We know by Gauss's Lemma that $(\frac qp) = (-1)^v$ where $$v = \#\left\{1 \le a \le \frac{p-1}{2} \mid aq \equiv k_a \pmod{12}),\space -p/2 &lt; k_a &lt; 0\right\}$$  And we also know that multiples of $(\dfrac{p-1}{2}) \times 3$ must be between $0$ and $p/2$, between $p/2$ and $p$, or between $p$ and $3p/2$.   Representing $p$ as $12k + r$ seems like the next step, but I'm not sure where to go from there.
17561
For which prime numbers p does the congruence $x^2 + x + 1 \equiv 0 \pmod{p}$ have solutions?  We've recently learnt about quadratic reciprocity in class, however I am not sure how to tackle this problem. I have tried starting with the $b^2-4ac$ (discriminant) but that hasn't really helped.
72123
Suppose $p$ is an odd prime. Prove that $r$, with $\gcd(r, p) = 1$, is a primitive root   modulo $p$ if and only if $r^{(p−1)/q}\not\equiv 1\pmod{p}$ for all prime divisors $q$ of $p − 1$.   The only if direction is trivial, I can multiply $q$ to get $(r^{(p-1)/q})^q\equiv1^q\pmod{p}$.  So, $r$ is a primitive root. How do I do the other direction? Can someone give me a hint or suggestion? Thanks
29122
The equation $y^2 = x^3 + k$ for $k = (4n-1)^3 - 4m^2$, with $m, n \in \mathbb{N}$ and no prime number that p is congruent to 1 modulo 4 divids m, doesn't have any answer and its proof can be obtained by using quadratic reciprocity law.  Do you know answers of this equation for two or three different values of $k$? In addition, do you know any reference about that?
116979
Given that $p&gt;3$ prove that $4p^2+1$ can be written as the sum of three distinct positive square numbers.   Plugging in $5$ I get $101=49+36+16=7^2+6^2+4^2$  I also know that all primes greater than $3$ can be written in the form $3k+1$ and $3k+2$ but plugging those values in I get:  $36k^2+24k+5$,  $36k^2+48k+17$ and the solution probably lies in arranging these numbers in such a way that we get the desired squares, but I can't come up with a combination ,or is my "idea" not even in the right direction?
41057
Find all primes $p&gt;2$ for which $x^2+x+1$ is irreducible in $\mathbb{F}_p[x]$   Attempt. Since $x^2+x+1$ is of degree 2, it is reducible iff it has a root in $\mathbb{F}_p$. It has a root in $\mathbb{F}_p$ iff $x^3-1=(x-1)(x^2+x+1)$ has a root other than 1. The latter has a root other than $1$ iff $\mathbb{F}_p$ has elements of order $3$, which happens iff $3\mid |\mathbb{F}_p^{\times}|=p-1$, which happens iff $p\equiv 1\bmod 3$. So $x^2+x+1$ is irreducible iff $p\not \equiv 1\bmod3 $.   I am getting a different answer than my professor, who has $p\equiv 1\bmod 3$. If I am incorrect, were am I going wrong?  Update. Now I am unsure about my own reasoning. What if $1$ is a root of $x^2+x+1$? I don't know why I am considering roots of $x^2+x+1$ other than $1$ when $1$ can be a root.
53865
For a prime number $p \neq 3$, do we have that$$p = x^2 + xy + y^2$$for some $x$, $y \in \mathbb{Z}$ if and only if$$p \equiv 1 \text{ mod }3?$$I suspect this is true from looking at the example$$7 = 2^2  + 2 \times 1 + 1^2.$$Here is a thought I have so far that might be helpful towards solving this: I know that $$\mathbb{Z}[(1 + \sqrt{-3})/2]$$is a PID. But I am not sure what do from here. Could anybody help?
34003
Are there any solutions to $x^{n}-1=p$ with p prime, integers $x,n&gt;1$ and $x$ not a power of $2$?  $x$ must be even. $n$ is odd since if $n=2m$ then $p=x^{n}-1=(x^{m}+1)(x^{m}-1)$ hence $p=x^{m}+1$ and $1=x^{m}-1$, which has solution $p=3$ given by $x=2^{2}$.   This generalises; for any solution to $x^{n}-c^{2}=p$, $n$ must be odd. And for any solution to $x^{n}-c=p$, $x$ is odd iff $c$ is odd unless $p=2$ and, if $a$ divides $c$ and $p$ does not divide $c$ then $x$ divides $c$.
120574
So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.  I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...  I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?
72373
Without strenuous arithmetic.   Is there a program I can download to do so?  What are the quadratic residues modulo $5^4$ or $5^5$?  Thanks!
26708
I'm currently learning how to find the inverse of a modulo with the Extended Euclid Algorithm and I stumbled upon a problem when finding an inverse when the $m&gt;p$ as for $m \equiv 1 \pmod{p}$  For example, in $$ 240x \equiv 1 \pmod{17} $$ what is the inverse? What are the steps to find it?
64493
I need to describe all numbers of the form $x^2 + 2y^2$.  So far I've reduced the problem to primes, and showed p=2 satisfies it. I've also shown that any primes mod 5 or 7 can't be a written in this form. How do I proceed to show that it holds for all primes mod 1 or 3? I think I'll need to use quadratic reciprocity somehow (only since it is the topic of the homework). I have a hunch the supplementary quadradic recprocity laws will be important.  Any help would be appreciated. Thanks!
65973
I have to compute $$2^{p-1} \mod p$$ and show by Fermat's little theorem that $p$ isn't prime. I know what the question is asking but I'm not sure how to reduce the exponent on $2^{p-1}$ to a more reasonable number. Does that make sense? Any general tips would be appreciated.
36462
As the title implies.  I'm proving it by contradiction and I've already ruled out $p \equiv 0$ and $2$ (mod $4$).  I'm having trouble proving that it cannot be $\equiv 3$ (mod $4$). Namely, if $p \equiv 3$ (mod 4), then $m^2 + n^2 \equiv 1$ or $3$ (mod $4$). I know that $m^2 + n^2$ cannot be congruent to $3$ (mod $4$); but what about $1$?  Thank you.
27657
According to the theorem, if  $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{p-1} =\frac{r}{q}$$  then we have to prove that $r\equiv0 \pmod{p^2}$.  (Given $p&gt;3$, otherwise $1+\dfrac{1}{2}=\dfrac{3}{2}$, $3 \not\equiv  0 \pmod 9$.)  I guess there's a $(\bmod p)$ solution for this, but I don't really get how to start it.
94191
Suppose that p ≡ 3 (mod 4)   and $r = \frac {p-1}2$  Show that $r! \equiv (−1)^k \pmod p$  where k is the number of non-quadratic residues modulo p which are smaller than $\frac p2$  I know from Wilson's Theorem for a prime p:  $(p−1)! \equiv −1 \pmod p$  I don't know where to start from ..   I managed to show that   $r! \equiv \pm 1 \pmod p$  any help or hint will be appreciated
25177
Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$  Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.  I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.