This question already has an answer here: solve $3x^2 + 6x +1 \equiv 0 \pmod {19}$ 2 answers

For any integer, $n$, show that $n^3 \equiv 0$ or $\pm 1(\mod 7)$. Use theory of congruences So I thought about a couple of ways to go with this. I thought about showing $7|n^3$ or $7|n^3\pm1$ to be true by letting $n=7k, 7k\pm1, 7k\pm2, 7k\pm3$ and proving it that way but that seemed like a […]

Determine the last two digits of $3^{3^{100}}$ This is one of the problems in the past exam my modern algebra course. I think I need to use euler-fermat theorem but can’t figure out how to use it for this problem. Can anyone help me out?

How to solve $x^3\equiv 10\pmod{990}$? It has 3 solutions: 10, 340, 670 (WolframAlpha).

How do we show $$n^{97}\equiv n\text{ mod }4501770$$ for all integer $n$? First of all, I thought I could use Fermat’s little theorem or Euler’s theorem, but I’m not sure if they are applicable here.

What should be the value of $n$ so that the number obtained after adding $1$ to $991$ times its square is itself a perfect square? Can you please give me a few hints on this topic with a few specific reasons?

My cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example, $$-48\mod{5} = 2$$ I kinda agree. But my grandpa insists that $$-48 \mod{5} = -3$$ Which is true? Why?

I need to show this by using Wilson’s Theorem, I don’t know how to start. Wilson’s theorem states $(n-1)! \equiv -1 \pmod n$

I’m having trouble solving this congruence: $$x^{114} \equiv 13 \pmod {29}.$$ I thought that it made sense to try to solve it using this idea: “Suppose you want to solve the congruence $ x^y \equiv a \pmod p$ (we will assume for the moment that $p$ is prime). Raise both sides to the power $z$ […]

I am asked to find all solutions of 1) $$x^2\equiv 1\bmod23$$ and to find all solutions of 2) $$x^{11}\equiv 1\bmod23,$$ justifying my work. I think 1) is $1$ and $22$ since if $p$ is an odd prime then $1$ has two distinct square roots modulo $p$, namely $1$ and $p-1$. However I am having difficulty […]

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