Can one find $x$ such that $a\mod p^n = x*b\mod p^n$?

Question

If given $n \in \mathbb{Z}^{+}$, $0 < a < p^n$ and $b \in \mathbb{Z}^{+}$ can I find $x$ such that

$$a\equiv xb\pmod {p^{n}}$$

If anyone has tips for where to further research the background material necessary to tackle this problem that would be sufficient.

It depends on $b$ and $a.$ If $b$ is relatively prime to $p^n$ then you can always solve it. If $p$ is a divisor of $b$ then $p$ must be a divisor of $a,$ or there is no solution. — Thomas Andrews, Sep 04 '19 at 05:49

score 0 · Accepted Answer · answered Sep 04 '19 at 06:19

Answer: if $a$ is a multiple of ${\rm gcd}(b,p^n)$, then yes; otherwise, no. Below, ${\rm gcd}$ means the greatest common divisor.

The tool you are missing is Bezout's identity:

For two nonzero integers $a,b$, there exists integers $x,y$ such that $$ax+by={\rm gcd}(a,b)$$

and its corollary

As $x,y$ both run through every integer, $ax+by$ runs through precisely the multiples of ${\rm gcd}(a,b)$. In other words, each $ax+by$ is a multiple of ${\rm gcd}(a,b)$; conversely, each multiple of ${\rm gcd}(a,b)$ has the form $ax+by$.

Let's return to your original question. Use the corollary you get $$a\equiv bx\pmod{p^n}\text{ has a solution}\\ \iff a-bx=p^ny\text{ for some integers }x,y\\ \iff a\text{ has the form }bx+p^ny\\ \iff a\text{ is a multiple of }{\rm gcd}(b,p^n)$$

Can one find $x$ such that $a\mod p^n = x*b\mod p^n$?

1 Answers1