Prove that $a^x\equiv a^{x\bmod(m-1)}\pmod m$ for m prime: I attempted to prove the cyclicity using Fermat's little theorem, that $a^p\equiv a\pmod{p} $ but didn't succeed. Could you please explain to me how to prove it?
3352
I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:   $13^{100} \bmod 7$ $7^{100} \bmod 13$   I've also heard of this formula:  $$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n $$  But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)  Any hints or pointers in the right direction would be great.
104857
I am trying to prove this statement which is equivalent to show $n^{91} = n^7 \pmod{ 364}$. By splitting modulus theorem $n^{91} = n^7 \bmod 91$ and $\bmod 4$. Then I don’t know what to do next... Can anyone help me?
6856
Here is a problem which i have not been able to do for quite sometime.   For $n \geq 2$, show that $n \nmid 2^{n}-1$.   I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have $2^{p-1} \equiv 1 \ (\text{mod} \ p)$, which actually says that for $p \mid 2^{p}-2$. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?
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.
123953
Can someone please show me how to prove this statement? $$\forall x\in\mathbb N (x &gt; 1\to\exists k\in\mathbb N\exists m \in\mathbb N (m \equiv 1 (\text{mod }2) \wedge x = 2^km))$$ I can only assume $m = 3$ and when $x &gt; 1$.
42349
I would like to show that $a^n \mid b^n$ implies $a \mid b$  I thought I could convert it to congruences and work backwards, but as far as I remember, $a \equiv b \pmod{m}$ implies $a^n \equiv b^n \pmod{m}$, not the opposite, unless $m$ is prime. Is that right?  So I am not sure how to approach this one. Any ideas?  Thanks!
110116
If  $e$ is a natural number, then this is true:  $$m^e \bmod\ n = (m\bmod\ n)^e\bmod\ n$$  This is often used when encrypting, especially with RSA, since one can avoid directly calculating $m^e$, which can be a very big number.  However, I haven't been able to find any documentation/proof for this conjecture, can anyone give a source or explain it?
21258
Prove that the multiplicative inverse of $a$ modulo $m$ exists if and only if $a$ and $m$ are coprime.  Can someone help me with this?
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.
80776
If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?  Here's what I've done:  $1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue, Also $(p-1)^3=(p-1) \pmod p$ implies  $$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p|(p-1)^3 -(p-1)$$ Thus  $p-1$ is a cubic residue $\mod p$. Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$.
36455
I'm having some trouble figuring out how to give a direct proof using the definitions of mod m and congruence modulo m, without using any theorem involving mod or congruence.  if $a, b, c$ &amp; $m$ are integers such that $m \geq 2$, $c &gt; 0$ and $a \equiv b \pmod{m}$, then $ac \equiv bc\pmod{mc}$
61119
If $|G| = pq$,  $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems?
16000
I was reading through a proof of the following proposition:  $$a\equiv b\!\!\pmod{m}\iff (a\bmod m) = (b\bmod m)$$  i.e. $\ a \equiv b \pmod{\!m} $ if and only if a and b leave the same remainder when divided by m   I came across a statement that I didn't quite understand. I boxed my area of confusion below:   I am confused because I expected the inequality to be $ 0 \leq r_1 - r_2 &lt; m$. How did they come to the conclusion that $ -m &lt; r_1 - r_2 &lt; m$  Any input is appreciated.
22479
Let $m,n$ be positive integers bigger than $1$. Show that $\mathbb{Z}_m \times \mathbb Z_n$ is cyclic if and only if $\gcd(m,n)=1$.    I have no idea on how to start. Anyone hints are much helpful.
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?
14039
Show that  $$a^2 \mathrm{\ mod \ } b = (a \mathrm{\ mod \ } b)^2 \mathrm{\ mod\ } b$$ for $ a, b \in \mathbb{Z}^+ $.  this was derived from an Informatics olympiad question.
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!