Congruency question: if $17x+11y \equiv 7 \pmod {29}$ and $13x+10y \equiv 8 \pmod {29}$, we need to find $x$ and $y$. There may be more than one answer. Not sure how to go about this; any help appreciated ðŸ™‚

EDITED QUESTION: I have this equation $$\left(x+\frac{2}{3}\right)\,mod\, \frac{13}{5}$$ and I want to find period after which its value repeats. It can be written as $$\left(\frac{3x+2}{3}\right)\,-\, \frac{13}{5}q = 0$$ I am equating it to some number (here I am equating it to zero to make the calculations simple), in order to find its period. And $q$ […]

Let $p$ be an odd prime. Suppose that $a$ is an odd integer and also $a$ is a primitive root modulo $p$. Show that a is also a primitive root modulo $2p$. Any hints will be appreciated. Thanks very much.

On Wikipedia, about Fermat quotients, it says: “Eisenstein discovered that the Fermat quotient with base $2$ could be expressed, $\mod p\ \ $ $p$ odd prime, in terms of the sum of the reciprocals of the numbers lying in the first half of the range ${1, p âˆ’ 1 }$”: $$ -2 \cdot \frac{2^{ p […]

I wish to show that $(p-2)! \equiv 1 \pmod p$ for a prime $p \ge 3$ using the fact that $(p-1)! \equiv -1 \pmod p$. (Deducing the latter is a later, and more advanced task.) We have that $(p-1)(p-2)! \equiv -1 \pmod p$. We have that $(p-1) \equiv (p-1)\pmod p$, but this does not help. […]

Is there a general way to find all values of $p$ such that the congruence $ax^2+bx+c \equiv 0 (\bmod p)$ have solution, we can assume that $ax^2+bx+c =0 $ have solution.

I am reading Notes of Mathematics for Computer Science(MIT 6.042J). And I’m stuck in the Modular Arithmetic Section which I mark with a RED LINE. I want to know what is trying to say. Is it trying to say that $(\operatorname{mod} 7)$ is neither associated with $29$ nor $15$?

I’ve checked a lot of the congruency posts and haven’t seen this one yet, so I’m going to ask it. If there is a related one, I’d be happy to see it. Let $x \equiv r\pmod{m}, x \equiv s\pmod{(m+1)}$. Prove $$x \equiv r(m+1)-sm \pmod{m(m+1)}$$ So with the given conditions, we know $$x=mk_1+r$$ Then $mk_1+r \equiv […]

I am asked to compute $\left(\frac{77}{257}\right)$ specifically using Euler’s criterion. I have manged to get the following: $$\left(\frac{77}{257}\right)\equiv77^{256/2} \bmod257$$ $$\equiv77^{128}\bmod257$$ I don’t know where to go from here as I don’t know the trick to simplify congruences that are big like this without typing it in on an online calculator. I believe it may be […]

$$p(p+1) \equiv -q(q+1) \bmod pq$$ Can this be reduced to an easier format?

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