Among some [permutation polynomials](http://en.wikipedia.org/wiki/Permutation_polynomial) I've been studying, $f(x)=(x+1)^n-x^n$ is one of the polynomials that I cannot grasp. 



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> **Question** : For a given **odd prime** $p$, how can we find **every positive integer** $n$ such that $f(x)=(x+1)^n-x^n$ is a permutation polynomial mod $p$ ?
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The answer seems $n=(p-1)m+2\ \ (m=0,1,2,\cdots)$, but I'm facing difficulty in proving that. This question has been asked previously on [math.SE](https://math.stackexchange.com/questions/1157778/finding-n-satisfying-1n-0n-2n-1n-cdots-pn-p-1n-equiv-0-1-cdot) without receiving any answers.


The followings are what I've got.


* $f(0)\equiv 1.$
* $f(p-1)\equiv -(-1)^n\Rightarrow \text{$n$ has to be even}\Rightarrow f(p-1)\equiv p-1$.
* For $n=(p-1)m+r$, $f(x)\equiv (x+1)^r-x^r$ because $a^{p-1}\equiv 1$ for $a$ which is coprime to $p$.
* $f\left(\frac{p-1}{2}\right)\equiv 0$.
* $f\left(\frac{p-1}{2}+a\right)+f\left(\frac{p-1}{2}-a\right)\equiv 0$ for any $a$.


I would like to know any relevant references as well.



Permutation polynomials mod $p$ of the form $(x+1)^n-x^n$
The expected answer is correct, it is a theorem by Norman Johnson, see [here.](http://dx.doi.org/10.1007/BF01223263) Note that $(X+1)^n-X^n$ is a permutation polynomial if and only if $(X+a)^n-X^n$ is a permutation polynomial for each fixed nonzero $a$. This latter condition is a special case of planarity: A function $f:K\to K$ on a field $K$ is called planar, if $x\mapsto f(x+a)-f(x)$ is bijective for every nonzero $a$.


So you ask for planar monomials on $\mathbb F\_p$ for odd primes $p$. Actually, it was later proved that any planar function on $\mathbb F\_p$ is given by a quadratic polynomial, see [this MO page](https://mathoverflow.net/questions/135949/a-question-on-maps-from-mathbbz-p-mathbbz-to-itself/136046) for more information and references.


