Let $(A,+,\cdot)$ be a ring such that there are $a,b \in A$ which satisfy $$(a+b)^2=a^2+b^2, \quad (a+b)^3=a^3+b^3$$ Prove that $(a+b)^n=a^n+b^n,$ for all positive integers $n.$
I have found the following solution, but I am not quite satisfied with it.
From the hypothesis we get $ab+ba=0$ and $ab^2+ba^2=0.$ We will prove the identity using induction. Suppose that it is true for $1,2,...,n-1, \: n \geq 4.$ We can write $$(a+b)^n=(a+b)^{n-1}(a+b)=(a^{n-1}+b^{n-1})(a+b)=a^n+a^{n-1}b+b^{n-1}a+b^n$$ It is left to prove that $a^{n-1}b+b^{n-1}a=0.$ We can write $$a^{n-1}b+b^{n-1}a=a^{n-2}ab+b^{n-2}ba=-a^{n-2}ba-b^{n-2}ab \quad (*)$$ But $(a+b)^{n-1}=(a+b)^{n-2}(a+b)=(a^{n-2}+b^{n-2})(a+b)=a^{n-1}+b^{n-1},$ so $$a^{n-2}b+b^{n-2}a=0 \Rightarrow b^{n-2}a=-a^{n-2}b$$ Plugging this back in $(*)$ gives $$a^{n-1}b+b^{n-1}a=-a^{n-2}ba+a^{n-2}b^2=a^{n-2}(-ba+b^2)=a^{n-3}ab(-a+b)$$ But $0=ab^2+ba^2=ab^2-aba=ab(b-a),$ so $$a^{n-1}b+b^{n-1}a=a^{n-3}\cdot 0 = 0$$ and this completes the indution.
Is there any other solution, maybe quicker or more beautiful?