Find possible primes p Suppose that $p$ is a prime and $n$ is a positive integer, and for every $a$ and $b$ that does not satisfy $a \equiv b \pmod p$, the congruence $-(a^n-b^n)+\binom{n}{1}(a^{n-1}-b^{n-1})+\binom{n}{2}(a^{n-2}-b^{n-2})+...+\binom{n}{n}(a^0-b^0) \equiv 0 \pmod p$ never holds. What are the possible values of $(n,p)$?
I've tried the binomial theorem but it makes stuff so compact and hard to manipulate... other theorems don't really work for all $n$...
 A: Some hints to get you started:


*

*First simplify the expression in the problem using the binomial theorem. Show that the problem can be reformulated as follows: With $f_n(x) = (1+x)^n - 2x^n$ the problem is to find all $(n,p)$ such that
$$a\not\equiv b\mod p \implies f_n(a) \not\equiv f_n(b) \mod p$$ or in other words find all $(n,p)$ such that  $$\{f_n(0),f_n(1),\ldots,f_n(p-1)\} = \{0,1,2\ldots,p-1\}$$ is a reduced residue system mod $p$.

*Next, I would always start by brute-force testing of the simplest cases $p=2,3,5$ with say $n=1,2,3,4,5$ to get an idea of how to proceed (if you don't already see this). If you do this then a simple pattern of allowed values for $n$ as a function of $p$ will reveal itself. This is what you should try to prove.

*What theorems to use? Note that $f_n$ is a sum of powers which just begs for Fermats little theorem. In order to apply it we should write $n$ in terms of $p-1$. To do this note that we can write a general $n$ on the form $n = i + (p-1)k$ with $0\leq i<p-1$ and $0\leq k$. Insert this into the expression for $f_n$ and apply Fermats theorem.

*Show that $i=1$ is always an allowed value (very easy).

*Finally try to show that other values for $i$ are not allowed (a bit harder). 
