## 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.

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## 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 $$x\equiv a \pmod{p_1p_2} ?$$

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