Prove that  $2^{6ℓ + 2}≡ 4 \pmod{18}$ for $0\leℓ ∈ ℤ$.
120507
What is the correct answer to this expression:    $26^{32}  \pmod  {12}$    When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$
59795
$p,q$  primes, $(pq,a)=1$ and $p-1|q-1$ then prove $a^{q-1}\equiv 1\pmod{pq}$   Got this in an exam today and had no idea how to do it.
54819
I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  So in all cases, I get the sum is congruent to $0$ regardless. Have I messed up, or is there a typo? Thanks.
11316
How can I solve $x^2 \equiv 19 \pmod {59}$?  I know that we can just try squaring numbers from 1 to 58 , but this is a very slow method, is not their a quicker one?
51735
I'm having difficult proving this.  As a hint the exercise to prove first, that if $a\lneqq \pm 1$ satisfies $a \equiv 3 \pmod4$, then exist $p$ prime, $p \equiv 3 \pmod 4$ such $p\mid4$. But I'm not really getting for what purpose can this be used.
71932
I am stuck with this exercise:     Show that $\forall s &gt;0 \quad \exists N &gt;0$ such that $$x^2 \equiv 1\pmod N$$ has more than $s$ solutions.   I think I cannot use the theory if quadratic residues, because the modulus is not prime, so maybe I must prove it with square roots or roots of unity, but I don't know how to approach it at all.  Thanks for any help.
26768
I was given this problem (I have to prove) and not sure if I use fermat's theorem  $24^{31} ≡ 23^{32} (mod 19)$  If I do use fermat's is this right:  I would do the LHS first:  $24^{18}·24^{13} ≡ 1·24·24^{12}$  RHS:  $23^{18}·23^{14} ≡ 1·23^{14}$  I am not sure where to from here or if it is correct thus far. I have looked at some other posts but they have not helped with this problem.
1672
I have had a go at this for a while now and can't seem to get anywhere.  $$2^{(2^{517})} \pmod {23}$$  I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.  Here is my attempt:  $$2^{(2^{517} \!\!\mod {22})} \equiv  2^{(2^{517})} \pmod {23}$$ Then proceed to do Chinese remainder theorem to find the exponent?
102996
Let there be integers $a,m,n$ such that $a=m^2=n^3$. Show that $a\equiv 0,1,28,36\pmod{63}$. I have established that $a$ must be a sixth power of an integer: $$a=\frac{a^3}{a^2} = \frac{m^6}{n^6} =\left (\frac{m}{n}\right )^6$$ $n^6\mid m^6$ implies $n\mid m$ so $\frac{m}{n}=:b\in\mathbb{Z}$. My objective is to show that $b\equiv 0,1,28,36\pmod{63}$. Then for every $f\in\mathbb{Z}[X]$: $$b\equiv r\pmod{n}\Longrightarrow f(b)\equiv f(r)\pmod{n} $$ (just pick $f(x)=x^6$). Trouble is, I have no idea how to show $b$'s congruences. How to proceed?    Alternatively, I could also state that $a = 63t+r, 0\leq r&lt;63$, but I don't know how to tie in the fact that $a$ is a perfect cube/square at the same time. Advice?
63331
So I know that if $(a,n)=1$, then $a^{\phi(n)}≡1\pmod{n}$ by Euler's Theorem. I also know that $d \mid a^k$ and obviously $d \mid n$. I know I probably need to use this in the proof, but I just don't know how to get started.
44076
How should one prove that $\binom n{0} + \binom n{3} + \binom n{6} +... +\binom n{3k} = \dfrac{2^n-2}{3}$ for $n$ odd and $\binom n{0} + \binom n{3} + \binom n{6} +... +\binom n{3k} = \dfrac{2^n+2}{3}$ for $n$ even, where $k=\lfloor \dfrac{n}{3}\rfloor $?
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?
14858
How do you prove this?  $$\left(n-1\right)^2\mid\left(n^k-1\right)\Longleftrightarrow\left(n-1\right)\mid k$$
64390
Let $p&gt; 3$ a prime number and $a, b \in \mathbb{Z}$ arbitrary.  Prove that $6p \mid (ab^{p} - ba^{p})$.
36599
How can we prove that every odd perfect square is congruent to $1$ modulo $8$?
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?
65965
Prove that:  (i)$5&lt;5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$  (ii)$8&gt;8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$  (iii)$n&gt;n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$  Can raising both sides to exponent $12$ help