# $c_n=a^n-b^n$, $S=\{c_1,c_2,...\}$ is finite then $0 \in S$

Statement:

Let $$(R,+,\cdot)$$ be a ring that has at least two invertible elements. Let $$a,b$$ be two invertible elements such that $$ord(a)=ord(b)=+\infty$$ Let consider the following sequence $$c_n=a^n-b^n$$ Let $$S=\{c_1,c_2,...,c_m,...\}$$ a set of elements from $$R$$.

Prove that if $$S=\{c_1,c_2,...\}$$ is finite then $$0 \in S$$

Attempt:

I obtained that:

$$a^n-b^n=c_n$$ then $$a^{n+1}-ab^n=ac_n$$ and $$a^{n+1}-b^{n+1}=c_{n+1}$$ then $$(b-a)b^{n}=ac_n-c_{n+1}$$ Since $$S$$ is finite then $$M=\{ac_n-c_{n+1},n \in \mathbb Z_+\}$$ is finite.

So $$\{(b-a)b^n,n \in \mathbb Z_+\}$$ is finite. Then there exists $$i$$ and $$j$$ such that $$(b-a)b^i=(b-a)b^j$$

How should I continue with this. I would highly appreciate somebody's help.

• Is there a connection to this question you had before? May 14 at 18:14
• Is the ring assumed to be commutative? May 14 at 21:46
• No, the ring is not commutative. May 15 at 6:50

Modified to NOT assume $$R$$ is commutative

Continuing from where you left off, you could use your line of reasoning to show the following: For any positive integer $$l$$ [not just $$l=1$$], there is a positive integer $$k_l$$ such that $$(a^{k_l}-1)c_l = 0.$$ [Indeed for each such $$l$$ there exists distinct integers $$i$$ and $$j$$ such that $$a^ic_l = a^jc_l$$, and assume $$j >i$$. Then $$k_l=j-i$$ works.] Furthermore, $$(a^k-1)c_l =0$$ for any $$k$$ that is a multiple of $$k_l$$, because even in a noncommutative ring, $$(a^{C_1}-1)$$ and $$(a^{C_2}-1)$$ commute for any two integers $$C_1$$,$$C_2$$. So let $$T$$ be a finite set of integers $$l$$ such that for all $$m \in \mathbb{N}$$ there is an $$l \in T$$ such that $$c_m=c_l$$, and set $$k = \prod_{l \in T} k_l.$$ Then $$(a^k-1)c_m = 0 \ \forall m \in \mathbb{N}.$$

In fact, for any multiple $$K$$ of $$k$$, it follows that $$(a^K-1)c_m =0 \ \forall m \in \mathbb{N}.$$ Thus we conclude with the following:

Claim 1 There is a positive integer $$k \in \mathbb{N}$$ such that the equation $$(a^K-1)c_m=0 \ \forall m \in \mathbb{N}$$ holds for all multiples $$K$$ of $$k$$.

We also make the following claim:

Claim 2 There exists a multiple $$K$$ of $$k$$ and a positive integer $$n$$ such that the equation $$c_{n+K}=c_n$$ holds.

Proof: Let $$N$$ be the number of distinct $$c_l$$; $$l \in \mathbb{N}$$. Then for any positive integer $$m$$, at least two of $$c_m, c_{m+jk}$$; $$j=1,2,\ldots, N$$; must be the same by the Pigeonhole Principle. ■

So, using Claims 1 and 2, let $$K$$ be a multiple of $$k$$, where $$k$$ is as in Claim 1, and let $$n$$ be an integer such that the equation $$c_n=c_{n+K}$$ holds.

Then simple algebra gives $$a^Kc_n +a^Kb^n -b^{n+K} = c_{n+K}.$$ plugging into this the equation $$c_n=c_{n+K}$$ and rearranging gives $$(a^K-1)c_{n+K} + c_Kb^n = 0.$$ But, as $$K$$ is a multiple of $$k$$ as in Claim 1, the equation $$(a^K-1)c_{n+K}=0$$ holds, so we are left with the equation $$c_Kb^n = 0.$$ As $$b^n$$ is invertible, it follows that the equation $$c_K=0$$ must hold. ■

• For the equation $$a^Kc_n +a^Kb^n -b^{n+K} = c_{n+K}$$ can you provide a little more detail? May 16 at 8:51
• $$a^Kc_n +a^Kb^n-b^{n+K}$$ $$= a^K(a^n-b^n)+a^Kb^n-b^{n+k}$$ $$= (a^{K+n}-a^Kb^n)+a^kb^n-b^{n+K}$$ $$=a^{K+n} -a^Kb^n+a^kb^n -b^{n+K}$$ $$=a^{n+K}-b^{n+K}$$ $$= c_{n+K}$$.
– Mike
May 16 at 15:27
• @quasi here it is in the above comment
– Mike
May 16 at 15:41
• Got it. Thanks.${}{}{}{}{}{}{}$ May 17 at 2:46