Questions tagged [congruences]

For questions about congruences in modular arithmetic, concerning for example the chinese remainder theorem, Fermat's little theorem or Euler's totient theorem.

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Method to Solve Linear Congruences

Can someone walk through how to solve $17x \equiv 7 \pmod{35}$? I'm having a lot of trouble with this and finding multiplicative inverses. I tried $\mathbf35 = 5(7) + 0$, but I'm not sure what to do since the remainder is 0
John
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Find all solutions of the equation $x^2+x=0$ in $\mathbb Z_6$

As in the title, I'm trying to find all the possible solutions to the equation: $x^2+x=0$ in $\mathbb Z_6$. I know that the solution for the congruence equation is valid only if $m |(b-a)$. I assume that the equation is already in that form, so the…
user306691
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If $y\equiv 1 mod 4$ then $y^2 \equiv 1 mod 8$

Given $$ y\equiv 1\ mod \ 4 $$ How can i proove that $$ y^2\equiv1\ mod\ 8 $$
greedsin
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Prove that $5a^2 \equiv k \bmod 12$ where $k \in \{0,5,8,9\}$.

Prove that $5a^2 \equiv k \bmod {12}$ where $k \in \{0,5,8,9\}$. Hence show that the equation $24x^7 + 5y^2 = 15$ has no integer solutions. I think I need to evaluate each case with the $k$ values first but not sure how...
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Understanding issue on demonstrating the congruences theorem

I've been working on congruences these days and figured out the core concept behind this notion. However I fail to understand a part of the demonstration: Part 1 was about proving that if $a\equiv b\pmod n$ then $n$ divides $a-b$. I got it, it was…
Erulk
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Is this correct? Is there a faster way to demonstrate it?

here are the instructions: 1) Prove that if $n$ is an even natural integer, then $n^2\equiv 0\pmod4$ 2) Prove that if $n$ is an odd natural integer, then $n^2\equiv 1\pmod4$ 1) If $n$ is even then it equals to $2k$ where $k \in…
Erulk
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What is $2^7 + 3^8 \bmod 11$?

What is $2^7 + 3^8 \bmod 11$? What is the most important thing that i should think of first when am faced with such kind of a problem. My tutor was like $2^7 = (2^5)(2^2)$. But why not raise it to different powers like 4 and 3 instead?
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Solve the system of congruences help?

I am really struggling with how to solve systems of congruences and I have a problem I need to solve as well as some attempt to solve it so any additional help would be so greatly appreciated! Equations: x = 3 mod 4 x = 4 mod 5 x = 6 mod 7 I…
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Equivalence of congruences - why are these congruences equivalent?

I'm reading a solution of the following congruence: $x^{59} \equiv 604 \mod 2013$. It says that it is equivalent to the following system of congruences: $$\begin{cases} x^{59} \equiv 604 \mod 3 \\ x^{59} \equiv 604 \mod 11\\ x^{59} \equiv 604 \mod…
qiubit
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Solve linear congruence: $ax + b = y \; (mod \; m)$

I am trying to solve $ax + b = y \; (mod \; m)$ for x, where $a,b,y,m$ are known values. This corresponds to running a linear congruential generator in reverse for one iteration. I am happy to assume that $gcd(a,m) = 1$, which according to this…
user154230
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Consider the system of Congruences $x \equiv a \pmod{m}$ and $x \equiv b \pmod{n}$

I want to prove that this system has a unique solution $\pmod{\frac{mn}{g}}$, where $g= \gcd(m,n)$ provided that $g \mid b-a$ Here is my attempt. From $x \equiv a \pmod{m}$ we know that $m \mid x-a$ and so there must exists an integer $k$ such that…
alkabary
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Modulus equation

I never seem to be able to understand this subject. "Give an expression for x when x(mod n) = x(mod m), while n and m are positive integers and n < m." My attempt: x(mod n) => x = an + c => c = x - an x(mod m) => x = bm + y => y = x - bm x(mod n)…
Arcthor
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Is it possible to get nontrivial $n$ such that we can find BOTH $n \pmod p$ and $\log{(n)} \pmod p$?

If we are looking for a value $n \equiv v \pmod p$ or $n \equiv v_r + v_i i \pmod p$, where $v_r+v_i\cdot i$ is a complex number modulo $p$, is it ever possible to have a situation where we can find both $n$ and $\log{(n)}$ modulo $p$? I know that,…
Matt Groff
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Getting the General Solution of Linear Congruence

$4x \equiv 12 \pmod {26}$ I have this equation and I understand that it has two solution via $\text{gcd}(26, 4)$. One of the answers is $x\equiv3$, which I can get by multiplying both sides by inverse of $4$. But the other answer which is…
raz789
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Prove that for each odd $m> 2$, is true that $\displaystyle \sum_{k=1}^{m-1} k \equiv 0 \pmod m$

please, know somebody solution for this argument? Prove, that for each odd $m > 2$, it is true that $$\sum_{k=1}^{m-1} k \equiv 0 \pmod m$$ Thanks for yours answers!
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