I have to find the number of solution of this congruential equation $x^k\equiv 1 (mod\space p)$. Surely I know that $x^{p-1}\equiv 1 (mod\space p)$ so, $x^d \equiv 1$ where $d$ is the MCD between $(p-1,k)$. Now I have to count how many divisor of $d$ there are, but I don't know how.
122878
I wish to find the number of solutions to the congruence $x^n \equiv 1 \mod p$ where $n$ is any natural number, $p$ prime. I tried using primitive roots and Fermat but I don't know where to go from here.
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.
114504
This is another one of the number theory problems I've been struggling with as of late (hopefully I'm not posting too many questions at once!).     Let $n$ be a positive integer and let $p=2^n+1$ be a prime number.   Prove that every quadratic nonresidue modulo $p$ is a primitive root   modulo $p$.   Suppose $a$ is a quadratic nonresidue and $r$ a primitive root muodulo $p$. Then there exists no $x$ such that $x^2 \equiv a \mod p$. We must determine the order of $a$. We know that $a$ must be congruent to some power of $r$, say $m$, since $\{r^1,r^2,\ldots,r^{2^n}\}$ is a reduced residue system modulo $p$. Suppose that $\mathrm{ord}_p a=b$. Then $a^b \equiv r^{mb} \equiv 1 \mod p$. We must show that $b=2^n$ since $\phi(p)=2n$. But $r$ is a primitive root and so $mb \equiv 0 \mod{2^n}$ (or $\equiv 2^n \mod{2^n}$). Hence $mb = 2^n k$. How can I finish it?
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.
20254
I'm having some troubles understanding the solution (going from step $2$ to step $3$) of this linear congruence equation. Here's the problem with the solution:   Solve the congruence $2x + 11 \equiv 7 \pmod 3$.   First, we reduce all the coefficients $\pmod 3$:  $$2x + 2 \equiv 1 \pmod 3. \tag{1}$$  Next, we add $1$ to both sides, using the fact that $2 + 1 \equiv 0 \pmod 3$:  $$2x \equiv 2 \pmod 3. \tag{2}$$  Finally, we multiply both sides by $2$, using the fact that $2 · 2 = 4 \equiv 1 \pmod 3$:  $$x \equiv 1 \pmod 3. \tag{3}$$  That is, any number in the set $\{\ldots, −5, −2, 1, 4, \ldots\}$ will solve the original congruence.    Like I said my problem is the step between $(2)$ and $(3)$. If I multiply both sides of $(2)$ by $2$ I get:  $$4x \equiv 4 \pmod 3.$$  Now, I understand that $4$ is congruent to $1 \pmod 3$ so the RHS "simplifies" to $1 \pmod 3$ but I don't understand why the LHS "simplifies" to $x$. Can someone help me with this?
106183
Find the number of solutions of $x^2=x\mod m$ for any integer $m$.  I was thinking about bringing the $x$ over to the other side and somehow apply chinese remainder thm.
98729
I've read through the elementary proof of why there exists a solution $x$ to $x^2\equiv -1\pmod p$ iff $p\equiv 1 \pmod 4$ for $p$ an odd prime. Is there a group theory generalization for this fact as well?
49397
How to solve $ 13x \equiv 1 ~ (\text{mod} ~ 17) $?   Please give me some ideas. Thank you.
117660
An exercise in number theory led to me to the following problem:     Find all solutions $(p,n,q,m)$ of the following equation:   $$\sum_{k=0}^n p^k = \sum_{h=0}^m q^h,$$   where $p&lt;q$ are distinct primes, and $1 \le m &lt; n$ are indeterminates.   Numerical evidence gives me the only solution $$(p,n,q,m)=(2,4,5,2).$$  There might be no other solution: I have no idea on how to show it. For those interested of the source of this equation, here it comes the interesting exercise in number theory.     Find all numbers $A$ such that the sum of divisors of $A$ divisible by   $5$ equals the sum of divisors of $A$ divisible by $2$: $$\sum_{2|d|A} &gt; d = \sum_{5|e|A} e. $$   Clearly this is equivalent to the condition that the sum of divisors of $A$ NOT divisible by $5$ equals the sum of divisors of $A$ NOT divisible by $2$. Let's factorize $A= 2^n \cdot 5^m \cdot w$, with $n,m \ge 0$ and $\gcd(w,10)=1$. Without loss fo generality we can consider $n,m\neq 0$. Then we get $$\sum_{d|2^nw} d = \sum_{e|5^mw} e$$ which is equivalent (using multiplicativity of the sum-of-divisors-function) to $$\sum_{k=0}^n 2^k = \sum_{h=0}^m 5^h$$ Giving us $A=2^4 \cdot 5^2 \cdot w=400w$ (with $\gcd(w,10)=1$). Thus, I am looking for simple generalizations with arbirary primes $p \neq q$.
30745
Let $p$ be a prime. Show that for all integers $x$, $x^2\equiv 1 (\text{mod} \ p) $ if and only if $x \equiv 1 (\text{mod} \ p)$ or $ x \equiv p-1 \ (\text{mod} \ p ). $    I know that I have to prove this both forward and backward. I'm not really sure where to begin
61377
I need to solve this equation:  $x^2 = 1$ in the ring $\mathbb{Z}/n \mathbb{Z}$.  I know  that  $(n - 1)(n - 1) \equiv 1 \pmod n$, in general $(n-a)^2 \equiv a^2 \pmod n$.  I also know that for $p$ prime all elements of $\mathbb{Z}_p$ are invertible and here $1$ and $p-1$ are their own inverses.  Could you tell me what other solutions there are to this equation?
57988
I was working on number theory homework, and this problem was the second part to  problem.  Let $a,b,c,d,m$ be constants and $n$ and $k$ be coprime such that there are infinitely many values of $x$ that satisfy  $$ax+b=mn$$ $$cx+d=mk$$  Find the values of $ad-bc$ such that the above is true.   Now, according to the answer suggested above, these equations are true if $n \equiv b/m\pmod a,$ and $k \equiv d/m\pmod c,$ but I can't really find any insight on how to find the values of $x$ that would help with this part of the problem.
77523
I've been asked to show that:  $n(n+1)(2n+1) \equiv 0 \pmod 6$  I found in a previous question that:  $n(n+1)$ was divisible by $2$ and resulted in an even number e.g  $n(n+1) \equiv 0 \pmod 2$  so I figured I needed to find:  $(2n+1) \equiv 0 \pmod 3$ in order to complete  $n(n+1)(2n+1) \equiv 0 \pmod 6$  but I am unsure on how to find $(2n+1) \equiv 0 \pmod 3$  Is this the right way to find the mod 6 and if so could you tell me how I could find  $(2n+1) \equiv 0 \pmod 3?$
57117
Find out all solutions of the congruence $x^2 \equiv29 \mod 5$.  [Hint:Find the solutions of this congruence $\mod 5$ , $\mod 7$ , and $\mod 7$ , and then use the Chinese Remainder Theorem.]
19988
What is the meaning of $$ \\\frac{m}{n}\equiv k\pmod l $$ and $$ \\\frac{m}{n} \mod l $$ $m, n, k, l \in\mathbb N$ and $\gcd(m, n)=1$.
80798
Let $x_1, x_2, x_3, x_4$ be  nonnegative  integers.      (a) Find the number of solutions to the following equation:                $$ x_1 + x_2 + x_3 + x_4 = 14 $$   I got $17 \choose 3$ for this. Is that correct?     (b) Find the number of solutions if we add the restriction that $x_i \le 6$ for        $1 \le i \le 4$
94934
Solve the congruence   $$4x\equiv16\mod{26}.$$  How do I find the solution to this? I have tried by the euclidean algorithm but the gcd is not $1$ so it doesn't work.  $$\begin{align} 26&amp;=&amp;6&amp;\times4&amp;+2\\ 4&amp;=&amp;2&amp;\times2&amp;+0 \end{align}$$  Note: I understand we can see that $4$ and $17$ are solutions but how would I work this out algorithmically?