# Tagged Questions

Modular arithmetic (clock arithmetic) is a system of integer arithmetic based on the congruence relation $a \equiv b \pmod{n}$ which means that $n$ divides $b-a$.

142 views

### Prove: “The cube of any number not a multiple of 7, will equal one more or one less than a multiple of 7”

Yeah so I'm kind of stuck on this problem, and I have two questions. 1. Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$...
627 views

18 views

### Generate Sieve of Eratosthenes without “sieve” (generate prime set in interval)

How do I generate a list of primes based on the Sieve of Eratosthenes? I mean by excluding the divisible numbers beforehand, which is tricky for large numbers. I am an number theory amateur, but was ...
306 views
+100

### Conjecture about primes and the factorial

Below $0\notin\mathbb N$. Further corrected conjecture: For all prime numbers $p>5$ there exist a prime number $q<p$ such that $q\equiv m!\!\pmod p$, $2<m<p$. or Given a prime ...
3k views

### How do I compute $a^b\,\bmod c$ by hand?

How do I efficiently compute $a^b\,\bmod c$: When $b$ is huge, for instance $5^{844325}\,\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, ...
90 views

56 views

### If the sum of two $p$th powers is divisible by $p$, then it is divisible by $p^2$

If $p > 2$ is a prime and $p | (x^p + y^p)$, then show that $p^2 | (x^p + y^p)$ I have been stuck on this problem for a while now. (Though my textbook is prone to mistakes so the original ...
29 views

### What is (a mod n) mod n?

I have an equation such as (a + b) mod n which is nothing but (a mod n + b mod n) mod n according to this. Now, I know that b mod n is 0 which results in (a mod n) mod n. Is this equivalent to a ...
Let $d$ and $n$ be coprime. What is the smallest positive solution for x in the equation: $$d^x \equiv 1 \mod n$$ This value must depend on both $d$ and $n$. We know that the maximum value for it is ...