1

This notation is found in Concrete Introduction to Higher Algebra.

Here is my method:

enter image description here

For something like $[3]_{11}[x]_{11}^2=[4]_{11}$ I've just been using C++ code like this:

#include <iostream>

using namespace std;

int main()

{

for(int j = 1 ; j<10 ; j++)
{
cout<< "{...,";
for(int i = -5 ; i<=10 ; i++)
{
cout << 11*i+2*j*j << ",";
}
cout << "...}";
cout << endl << endl;
}

return 0;

}

Andrés E. Caicedo
  • 79,201
Yosef Qian
  • 477

4 Answers4

4

Algorithmic method

I would use the Euclid-Wallis Algorithm to solve $17x+4y=1$: $$ \begin{array}{r} &&4&4\\\hline 1&0&1&-4\\ 0&1&-4&17\\ 17&4&1&0\\ \end{array} $$ This says that $17(1)+4(-4)=1$. This means that $-4\equiv13\pmod{17}$ is the inverse of $4$ mod $17$. To get $2$, just double to get $26\equiv9\pmod{17}$. Thus, $$ 4\cdot9\equiv2\pmod{17} $$

Non-algorithmic method

We're trying to solve $4x\equiv2\pmod{17}$ which would be $x=\frac12$ if we could divide by $2$. However, we can divide $1\equiv18\pmod{17}$ by $2$ to get $x=9$. Care must be taken; we can only divide by numbers which are relatively prime to the modulus.

Community
  • 1
robjohn
  • 345,667
4

Note that $4\cdot 4=16\equiv -1\pmod{17}$, hence $1/4=-4\pmod{17}$ (meaning nothing else but $(-4)\cdot 4\equiv 1$), so we have to solve $$\begin{align} 4x&\equiv 2 &\pmod{17} \\ x&\equiv (-4)\cdot 2&\pmod{17} \end{align}$$ That is, $x\equiv -8\equiv 9\pmod{17}$ is the solution.

Berci
  • 90,745
2

You're looking at $$4x\equiv 2\mod 17$$

Since $(4,17)=1$; this has a unique solution, and it can be found by finding the inverse for $4$ modulo 17. But $4\times 4 =16\equiv -1\mod 17$, so $-4\equiv 13\mod 17$ is the inverse of $4$. It follows that $$13\cdot 4x=13\cdot 2\mod 17$$

$$x=13\cdot 2\mod 17$$

$$x=9\mod 17$$

Pedro
  • 122,002
  • (+1) but as a lazy mathematician, I object to writing $-4 \equiv 13$, then multiplying $13$ by $2$ and finally reducing modulo $17$. Computing $-4 \cdot 2$ and then reducing modulo $17$ is much easier :-) – TMM May 10 '13 at 00:31
  • @TMM Whatever floats your boat. =D – Pedro May 10 '13 at 00:34
0

If you knew the inverse of $4 \pmod{17}$, you could multiply it on each side of your equation and compute $x$. Let $a$ be the inverse of $4 \pmod{17}$, that is, $4*a\equiv1\pmod{17}$. To find $a$, find a number that's congruent to $1\pmod{17}$ and divisible by 4, then divide it by 4. So start with 1 and add 17's until you get a multiple of 4, then divide by 4. You can do this calculation more easily modulo 4, noticing that $17\equiv1\pmod{4}$, so $1+3*17=52$ is divisible by 4, giving $a=13$. Now multiply each side of your equation by 13, to find $x\equiv 2*13 \pmod{17}\equiv26 \pmod{17} \equiv 9 \pmod{17}$.

Josh B.
  • 3,376