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.

Expected value of offer he will accept, given he will accept for X>$k

Jack is selling his house, and has decided to accept the first offer exceeding $ k. Assume that successive offers are independent random variables with common distribution Exp(λ). Find the expected value of the offer he will accept.

I know by indepence P(X1≤r)=P(X2≤r)=P(Xi≤r)=F(r).

P(Xn offer is accepted)=P(X1≤r)P(X2≤r)…P(Xn−1≤r)P(Xn>r)=F(r)n−1(1−F(r)). Where p=e^-λr, P(N=n)=(1−p)^(n−1)p. I then found E(N)=∑nP(N=n) and working through gets me E(N)=1/p.

As P(N=n) is geometric,P(X)=1/E(N)=e^-λr.

I have seen that E(X>k)=∫xP(x) dx/∫P(x) dx from k to infinity, but I can’t seem to reduce this.

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} ?$