In a ring where $(a-b)^2 = a - b$ for fixed $a,b$, then $(a-b)(a+b) = 1 \iff a^2 - b^2 = 1$.

Question

Let $(A,+,\cdot)$ be a ring where $(a-b)^2 = a - b$ for fixed $a,b$. Then $$(a-b)(a+b) = 1 \iff a^2 - b^2 = 1.$$

I was able to prove one implication:

Proof ($"\Rightarrow")$

We have that $(a-b)^2(a+b) = a-b = (a-b)(a+b) = 1$ thus $a-b = 1$ so $a+b=1$. Then $a + b - (a - b) = 1 - 1 = 0 = 2b$;

But since $a-b = 1$ we get $a = b + 1$ so $a^2 = b^2 + 2b + 1 = b^2 + 1$ so

$$a^2 - b^2 = 1$$.

Any tips for the other implication?

Answer 1

Idempotents are always a joy to work with. For the converse implication denote $a-b=e$ and notice that $e$ is an idempotent by hypothesis; substituting $a=b+e$ in the relation $a^2=b^2+1_A$ leads (after a few simple calculations and cancellations) to $e+be+eb=1_A$ and hence to $$eb+be=1_A-e \tag{1}$$

Multiplying relation (1) by $e$ on the left as well as on the right yields

$$eb+ebe=ebe+be=0_A \tag{2}$$

and hence to $eb=be=-ebe$.

Since $b$ and $e$ commute, $b$ and $a=b+e$ must also commute and thus one can factor the difference of squares $$1_A=a^2-b^2=(a-b)(a+b)$$