Sunday 26 March 2017

elementary number theory - Solving congruence system with no multiplicative inverse




I am trying to find a way of solving congruence systems of the form:



$$ b*x = a \quad mod \quad y $$




Where $b$ and $y$ are not prime to each other.



My current way of solving congruence systems where $b$ and $y$ are prime to each other is to find the multiplicative inverse of $b$ in $y$ and multiply $b$ (which will make it 1) and $a$ with this value.



Example:



$$ 13 \ x = 3 \ mod \ 17$$



I calculate the mul. inverse of 13 and multiply 3 and 13 times that value and thus I have solved this equation.




But I dont know how I can do that where $b$ and $y$ are not prime to each other.



Example:



$$3 x = 3 \ mod \ 9$$



How would I solve this ?


Answer



A congruence

$$ax\equiv b\pmod{n}\tag{1}$$
is soluble iff $\gcd(a,n)\mid b$. In this case (1) is equivalent to
$$\frac agx\equiv \frac bg\pmod{\frac ng}\tag{2}$$
where $g=\gcd(a,n)$. As $a/g$ is coprime to $n/g$ you may solve (2)
by multiplicative inverses if you like.


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