It is given that $a^n-1$ is divisible by $n$ for some natural numbers $a$ and $n$. Prove that the numbers $$a+1,a^2+2, \ldots a^n+n$$ all leave different remainders when divided by $n$.
I tried applying some modulo relations which proved unsuccessful. Some hints, please. Thanks.
The condition $a^n \equiv 1 \mod n$ motivates the introduction of $ord_n(a)$. Then we see that we can partition the $n$ numbers in groups which have specific modular properties. Induction seems convenient. Altogether, this motivates the following solution...
We are going to prove this using strong induction over $n$.
For $n=1$ it is clearly true.
Assume we proved it for $n=1,2,\dotsc,m-1$. Let's prove it for $n=m$.
Let $a$ be any number such that $m|a^m-1$. Denote $ord_m(a)=k$. We have $m=k \cdot r$ and $a^k \equiv 1 \mod m$. Also, since $ord_m(a)|\phi(m)$ by FLT we conclude $k \le \phi(m)<n$.
Denote $a^{i}+i=x_i$. Now fix any $s$ with $1 \le s \le k$ and look at the numbers $x_s,x_{k+s}, x_{2k+s}, \dotsc$.
Note that $x_{ik+s} \equiv a^s+ik+s \equiv x_s+ik \mod m$. Hence (since $k|m$) we know that the numbers $x_s, x_{k+s},x_{2k+s},\dotsc$ leave all the same residue $\mod k$.
Also since $x_{ck+s}-x_{dk+s} \equiv ck-dk \equiv (c-d)k \mod m$ and $r \not \mid c-d$ they all leave different residues $\mod n$.
Since there are only $r$ values of the form $ik+x_s \mod n$ we conclude that all these residues are indeed assumed by some $x_j$.
We can repeat the same argument for all $s$ from $1$ to $k$.
Now, assume that some residue $x \mod n$ is not assumed by any of the $x_j$. Then we look at $x \mod k$ and find that this residue is not assumed by any of $x_1,x_2,\dotsc,x_k$ (otherwise it would be assumed by our above argument).
But our induction hypothesis states that the numbers $x_1,x_2,\dotsc,x_k$ assume all residues $\mod k$ and thus the assumption must be wrong.
Hence we have established the induction step and thus proved the original claim.
