I need to prove that $x^2\equiv -1\pmod p$ if $p=4n+1$.  ($p$ is prime of course...)  I need to use Wilson theorem.
4827
I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:     Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.   I appreciate any help.  Thanks.
23623
Let $p=4k+1$ be a prime number.  I need to  prove that $p$ divides $k^k-1$.  Could anyone give me some hint?  Thank
36335
I need to show that if $(n-2)! \equiv 1 \pmod n$ then $n$ is prime. I think that if $n$ is composite, then we'll have every factor of $n$ in $(n-2)!$, and it would yield that $(n-2)! \equiv 0 \pmod n$. However, I didn't use the fact that it specifically congruent to $1 \bmod n$, so I think I'm getting something fundamental wrong. Is my solution correct? Why do we demand congruence to $1 \bmod n$?
46776
I really need some help at this problem:     Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n  \mid  p-1$ and $p \mid  n^3-1$. Prove that $4p-3$ is a square.   So $p \mid (n-1)(n^2+n+1)$  What if $p \mid n-1$?   Treating the cases wasn't too efficient. I was thinking about Fermat's theorem but it didn't helped really much.   A hint would be really apreciated. Thanks!
107659
Prove or disprove $$\dfrac{n^2 + 1}{4k + 3} \notin \mathbb{Z}, n, k \in \mathbb{N}^{+}$$  I know if $n^2 + 1$ is prime if and only if $n^2 + 1 \equiv 1 \pmod 4$.
27206
a, b are integers. p is prime. I want to prove: $(a+b)^{p} \equiv a^p + b^p \pmod p$  I know about Fermat's little theorem, but I still can't get it I know this is valid: $(a+b)^{p} \equiv a+b \pmod p$ but from there I don't know what to do.  Also I thought about $(a+b)^{p} = \sum_{k=0}^{p}\binom{p}{k}a^{k}b^{p-k}=\binom{p}{0}b^{p}+\sum_{k=1}^{p-1} \binom{p}{k} a^{k}b^{p-k}+\binom{p}{p}a^{p}=b^{p}+\sum_{k=1}^{p-1}\binom{p}{k}a^{k}b^{p-k}+a^{p}$ Any ideas?  Thanks!
38927
How can I show that $(n-1)!\equiv-1 \pmod{n}$ if and only if $n$ is prime? Thanks.
19013
I came across a claim that I found interesting, but can't seem to prove for some reason. I have the feeling it should be easy  a prime $p$ can be written in the form $p = a^2 -ab +b^2$ for some $a,b\in\mathbb{Z}$ if and only if $p\equiv 1\bmod{6}$
56575
how to prove $\forall p$ prime :   $\binom{2p}{p} \equiv 2 \pmod p$ we have:  $\binom{2p}{p} = \frac{2p (2p-1)(2p-3)...1}{p!p!}$ but how to continue?
18060
let q be an odd prime and p a prime dividing $1+n+\dots+n^{q-1}$. Prove that $p = 1 \mod q$ or $p = q$.  I don't actually know how to start, i thought of using fermat's little theorem or an application of eulers formula, but it didn't really give me any insights.   Any hints?
31892
Assume $p$ is prime and $p\ge 3$.  Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that.
32031
I have heard somewhere that for all primes $p$, for all $k$, there exist $x, y$ s.t. $x^2 + y^2\equiv k \pmod{p}$? I recall that the proof is very elementary, but I can't remember such a proof. How would one prove this? One way is to use Cauchy-Davenport, but I don't think that this is the simplest solution.
26965
Prove or disprove: For a prime number p, there is a positive integer n such that $p+1 $= $n^2$ if and only if $ p = 3.$  first im trying to proof if p is prime number there is positive integer n such that $p+1=n^2$ implies $p=3$  suppose there is positive n integer such that $p+1=n^2$ and prove that $p=3$. $p=n^2-1=(n+1)(n-1)$ since $p$ is prime $n-1=1$,$ n = 2 $ $p+1=4$ $p=3$ hence it is proved that $p=3$  now if $p=3$ implies there is positive integer n such that $p+1=n^2 $  $p=3$ and p is prime  $p+1=n^2$, $4=n^2$, $n2$ or $n=-2$ hence it is proved that there is positive integer $n$ such that $p+1=n^2 $  is my proof correct? and also for prime number can i write $n+1=1$ instead $n-1=1$?
56173
Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ for $p\neq q$ distinct primes.  Where can I start with this proof? It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.
54679
It is a theorem in elementary number theory that if $p$ is a prime and congruent to 1 mod 4, then it is the sum of two squares. Apparently there is a trick involving arithmetic in the gaussian integers that lets you prove this quickly. Can anyone explain it?
68539
Prove that for a prime $p&gt;2$, if $p \mid m^p+n^p$, prove $p^2\mid m^p+n^p$  From Fermats theorem I concluded $p \mid m+n$, so $p^2\mid (m+n)^p$. How do I proceed next? Any hints are welcomed.
35299
Proof that if $p$ is a prime odd and $k$ is a integer such that $1≤k≤p-1$ , then the binomial coefficient  $$\displaystyle \binom{p-1}{k}\equiv (-1)^k \mod p$$  This exercise was on a test and I could not do!!