I'm just a little stuck on this problem. I think the solution must incorporate something like Fermat's theorem, but I'm struggling with getting it exactly correct.   Show that if $p$ is a prime number and $a$ and $b$ are natural numbers, then   $$(a + b)^p \equiv a^p + b^p \pmod{p}$$
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!
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!
57250
What I have done is this:  $a\equiv b \pmod{2n}$,  $a=b+c\times2n$, for some $c$,  $a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,  $a^2-b^2=(b\times c+c^2n)\times4n$, then  $a^2\equiv b^2\pmod{2^2n}$.  I think that this is right: what I DON’T understand is how to generalize this to:  $a\equiv b\pmod{kn}\Rightarrow a^k\equiv b^k \pmod{k^2n}$.  Please give me a hint.
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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 $n-1=1$?
35054
Which of the following primes satisfy the congruence   $$a^{24}\equiv6a+2\pmod{13}$$  1) 41  2) 47  3) 67  4) 83  I am interested in Theorem statement, corollary, or Trick or Logic which solves this problem within one minute. Thank you in Advance
112039
Problem: show that if the prime $p = a^2 + 5b^2$ then $p \equiv 1 \textrm{ or } 9 \pmod{20}$.  Is there any other way to prove this other than to check each case $p = 1, \ldots, 19 \pmod{20}$ and show that only 1 and 9 work?
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?
54889
Sequence $a_0,a_1,a_2,...$ satisfies that $a_0=2,a_1=1,a_{n+1}=a_n+a_{n-1}$  Prove that if $p$ is a prime divisor of $a_{2k}-2$,then $p$ is also a prime divisor of $a_{2k+1}-1$    If $x_{1,2}={1\pm\sqrt{5}\over 2}$ then $a_k = x_1^k+x_2^k$ and $a_{2k}-2(-1)^k = (x_1^k-x_2^k)^2$.
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I need help with this question.     Let $a$ and $b$ be positive integers such that $a^{n}+n$ divides $b^{n}+n$ for every natural number $n$. Show that $a=b$.   Any help would be appreciated! Thanks!
117611
Let $p$ be a prime such that $a^p-1 \equiv 0 \pmod{p}$. Prove that $\gcd(a,p) = 1$.   We know from Fermat's Little Theorem that $a^{p-1}-1 \equiv 0 \pmod{p}$ if and only if $\gcd(a,p) = 1$, but how do we use this to solve the question?
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Let $p$ be a prime satisfying $p\equiv 3 \mod 4$. Let $a$ be a quadratic residue modulo $p$. Prove that the number $$b\equiv a^\frac{p+1}{4} \mod p$$ has the property that $b^2\equiv a \mod p$. (Hint: Write $\frac{p+1}{2}$ as $1+\frac{p-1}{2}$.) This gives an easy way to take square roots modulo $p$ for primes that are congruent to $3$ modulo $p$. \textit{Hint:} Write $\frac{p+1}{2}$ as $1+\frac{p-1}{2}$ and use Exercise $3.36$.) This gives an easy way to take square roots modulo $p$ for primes that are congruent to $3$ modulo $p$.  I assume that the proof comes directly from the proof of quadratic residues but I am not sure how.
38927
How can I show that $(n-1)!\equiv-1 \pmod{n}$ if and only if $n$ is prime? Thanks.
12941
The proof is a follows:  Let $p$ be a prime number and $x$ in Z. If $x^2\equiv1\pmod p$ then $x\equiv\pm1\pmod p$.  Proof: By assumption, $p | x^2 - 1 = (x + 1)(x - 1)$. Thus $p | x + 1$ or $p | x - 1$. This completes the proof.    However if $p | x + 1$ then $x$ is congruent to $-1 \mod p$. However why is $p$ congruent to $1 \mod p$ ??   I'm struggling seeing this, and been sitting for some hours now trying to figure it out - please help.
62474
I want to prove that if $p$ is a prime and $x,y \in \mathbb{Z}$, then $$(x+y)^p \equiv x^p+y^p \pmod{p}$$  So far I know that $$(x+y)^p = \sum_{k=0}^{p} \dbinom{p}k x^{p-k} y^k$$  A part of the above equation is supposed to cancel, I think, but I can't figure out a way how to make it cancel.
70953
I believe that this may require the use of Fermat's Little Theorem. I rewrote it as $(a+1)^p - a^p \equiv 1 \pmod{p}$ because the right-hand side looks similar to Fermat's Little Theorem, but I was unable to figure out how I can get the left-hand side to become $a^{(p-1)}$.
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.
82511
For a prime $p \equiv 1$ or $3$  (mod $8$), show that the equation $x^{2} + 2y^{2} = p$ has a solution.   I'd like to see an elementary proof.