Congruence mod $n+1$ of radix-$n$ integers

Question

Let $a$ and $b$ be distinct positive integers. Prove (or disprove) that if $a\equiv b\pmod{n+1}$, then the radix-$n$ representation of $a$ and $b$ differ in at least two digits.

Edit to add some more details:

I arose at this conjecture while trying to find a simple way to partition integers into a small number of sets in such a way that no two integers in a set differ by only a single digit. I have only verified for a relatively small number of cases, but it seems to work. It makes some intuitive sense to me for reasons that are tricky to verbalize.

As for an attempt at a proof, my initial thought was to observe that if $a$ and $b$ differ in only a single digit, then $a-b=cn^d$ for some integers $1\leq c<n$ and $d\geq0$. Then, if $a\equiv b\pmod{n+1}$, then $(n+1)\mid cn^d$. But from here I'm stuck.

Answer 1

Assume without loss of generality that $a>b$. If $a$ and $b$ differ in just one digit, then there must exist integers $c\in[1..n-1]$ and $e\geq0$ such that $a-b=c\,n^e$. Now, since $a\equiv b\bmod{(n+1)}$, it follows that $(n+1)\mid c\,n^e$. But $\gcd(n,n+1)=1$, so (using an easy generalization of Euclid's Lemma to arbitrary integers) this can only happen when $(n+1)\mid c$, which is clearly impossible due to the restriction that $c\in[1..n-1]$.

(For the claim that $\gcd(n,n+1)=1$: using the fact that $\forall x,y\in\mathbb{N},\gcd(x,y)=\gcd(x,y-x)$, it follows immediately that $\gcd(n,n+1)=\gcd(n,1)=1$.)