# Tag Info

Accepted

### Prove that $\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$ with $p$ being an odd prime

The result can be easily proved without using Bernoulli numbers. If $a$ and $b$ are integers not divisible by an odd prime $p$, then \begin{align}(ab)^{p-1}-1=&b^{p-1}(a^{p-1}-1)+(b^{p-1}-1) \\\...
• 14.6k
Accepted

### A good reference to the general Chinese Remainder Theorem

It seems that you are after this result which can be found, for example, as Theorem 3.12 in Gareth A. Jones, Josephine M. Jones: Elementary Number Theory, Springer-Verlag, London, 1998. Springer ...
• 4,628
Accepted

### Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$

Yes, this is true, as a consequence of an identity in a space of modular forms of weight $6$. The form $\eta(2z)^{12}$ is in this space; and $\sigma_5(n)$ for $n$ odd are the coefficients of the ...
• 77.8k
Accepted

### What did Yu Jianchun discover about Carmichael numbers?

Apparently it is an alternative proof of the infinitude of Carmichael numbers. The other proof mentioned in the articles ("done by academics 20 years ago") is: W. R. Alford, Andrew Granville, Carl ...
• 17.5k
Accepted

• 104k
Accepted

### For which values of $k$ is it known that there are infinitely many $n$, such that $2^{n+k}\equiv 1\pmod{n}$?

For any $k\geq 1$, there are infinitely many solutions of the congruence $2^{n+k}\equiv 1\pmod{n}$. To see this, observe first that there is always a solution $n\geq 1$ satisfying $n+k\geq 7$. Indeed, ...
• 101k