## What is the time complexity of determining whether a solution $x$ exists to $x^k \equiv c \pmod{N}$ if we know the factorization of $N$?

Suppose we are given an integer $$c$$ and positive integers $$k, N$$, with no further assumptions on relationships between these numbers. We are also given the prime factorization of $$N$$. These inputs are written in binary. What is the best known time complexity for determining whether there exists an integer $$x$$ such that $$x^k \equiv c \pmod{N}$$?

We are given the prime factorization of $$N$$ because this problem is thought to be hard on classical computers even for k = 2 if we do not know the factorization of $$N$$.

This question was inspired by this answer, where D.W. stated that the nonexistence of a solution to $$x^3 \equiv 5 \pmod{7}$$ can be checked by computing the modular exponentiation for $$x = 0,1,2,3,4,5,6$$, but that if the exponent had been 2 instead of 3, we could have used quadratic reciprocity instead. This lead to my discovery that there are a large number of other reciprocity laws, such as cubic reciprocity, quartic reciprocity, octic reciprocity, etc. with their own Wikipedia pages.

## Given $L$ and $D$ find $X, \text { such that } X * 10^L + D \equiv 0 \mod M$

Given $$L$$ and $$D$$ find $$X, \text { such that } X * 10^L + D \equiv 0 \mod M$$. Integer $$M$$ is given and it is the same for all calculations however we need to solve for $$X$$ for more different numbers. One important thing that we know is that $$\gcd(M, 10) = 1$$.

I rewrited the equation in this type: $$X * 10^L \equiv M – D \mod M$$. If $$M$$ was prime number we could just multiply $$M-D$$ by $$(10^L)^{M-2}$$. However $$M$$ might be arbitrary integer. How can we use the fact that $$\gcd(M, 10) = 1$$

## What is the difference between “$=$” and “$\equiv$”?

I was recently thinking about some of my past math classes, and depending on the context I recall my professors would sometimes use the “$$\equiv$$” symbol in places where I’d feel “$$=$$” to be more appropriate. For example, since this would often be the case in my classes on differential equations and Fourier series, we would have (for $$n \in \Bbb N, k \in \Bbb Z$$)

$$(-1)^{2n+1} \equiv -1$$ $$\sin(k\pi) \equiv 0$$

Is there a particular reason in this context why we would say “$$\equiv$$” instead of “$$=$$“? The latter feels more natural in this context, which makes me think that there’s some reason my professors would use the former.

I’m familiar with the notion of the “$$\equiv$$” symbol in the context of, say, elementary number theory (specifically modular arithmetic) where we might say

$$10 \equiv 1 \pmod 3$$

which isn’t saying “$$10$$ equals $$1$$“, just that “$$10$$ is like $$1$$ in this context.” But that doesn’t seem to fit the case as with the first two statements – because I don’t believe it is that $$(-1)^{2n+1}$$ is like $$-1$$, or that $$\sin(k \pi)$$ is like $$0$$, they are $$-1$$ and $$0$$ respectively.

Am I just mistaken on this latter fact? Is there something I’m missing? What, precisely, is the difference between the two notations?

## Prove that $3^{30} \equiv 1 + 17 \cdot 31 \pmod{31^{2}}$.

I guess this problem is easy, but I cannot solve it.

Prove that $$3^{30} \equiv 1 + 17 \cdot 31 \pmod{31^{2}}$$

Of course, I can solve the above problem by direct calculation, but I wanna know smarter solution.

I did the following calculation for example, but I was not able to solve this problem.

By Fermat’s little theorem,
$$3^{30} \equiv 1 + 17 \cdot 31 \pmod{31}.$$

$$1 + 17 \cdot 31 \equiv (1 + 31)^{17} \equiv 32^{17} \equiv 2^{85}\pmod{31^2}$$

## exist $x$ such that $x^k \equiv m$ mod $(p_1\cdot p_2) \Leftrightarrow$ exists $x_1,x_2$ : $x_1^k\equiv m(p_1)$ and $x_2^k\equiv m(p_2)$

exist $$x$$ such that $$x^k \equiv m$$ mod $$(p_1\cdot p_2) \Leftrightarrow$$ exists $$x_1,x_2$$ : $$x_1^k\equiv m(p_1)$$ and $$x_2^k\equiv m(p_2)$$

A first approach I took, was to use $$y\equiv m(p_1) , y\equiv m(p_2) \Leftrightarrow y\equiv m(p_1\cdot p_2)$$, then by assigning $$x^k=y$$ the problem comes to find whether $$y$$ has a $$k$$-order root in $$U(\mathbb{Z}_{p_1\cdot p_2})$$. How ever it doesn’t seem to simplify the problem.

A second approach I took was to use the fact which derived from CRT , that $$U(\mathbb{Z}_{p_1 \cdot p_2}) \cong U(\mathbb{Z}_{p_1}) \times U(\mathbb{Z}_{p_2})$$, In $$U(z_{p_i})$$ which are cyclic groups, there is a solution for $$x^k \equiv m(p_i) \Leftrightarrow m^{\frac{p_1-1}{gcd(k,p_1)}}=1 (p_i)$$. So assuming $$gcd(k,p_1) = 1$$ there are solutions for the equations $$x_1,x_2$$. But I am struggling to show that $$\pi^{-1}(x_1,x_2)$$ (when $$\pi$$ is the isomorphism from CRT), is a solution for $$x^k \equiv m (p_1p_2)$$.

So in case my second approach is correct, I would be glad for some help with showing $$\pi^{-1}(x_1,x_2)$$ is a solution, and also in case $$x$$ is a solution mod $$(p_1p_2)$$ then $$\pi_1(x) ,\pi_2(x)$$ are solutions mod $$p_1$$, $$p_2$$ respectively.

Also other approaches or ideas would be appreciated.

related question:

If $$x \equiv a \pmod {p_1}$$ and $$x\equiv a \pmod{p_2}$$, then is it true that $$x\equiv a \pmod{p_1p_2} ?$$