Showing the formal identity $(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$

Question

How to show this formal identity (or you can assume $|x|<1$)? $$(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$$

I can show that the latter is $$=1-x+x^3-x^4+x^6-x^7+\cdots$$ but how to show this is equal to the infinite product. I think it has something to do with residue of the exponent modulo $3$.

Answer 1

Notice that $$(1+x^n+x^{2n})(1-x^n+x^{2n}) = 1+x^{2n}+x^{4n}$$

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Now if $p_n$ is product of first $n$ brackets on LHS then $$p_n ={1+x^{2^n}+x^{2^{n+1}}\over 1+x+x^2}$$

Taking $\displaystyle \lim _{n\to \infty}p_n$ for $|x|<1$ we get the desired.

Answer 2

Notice that $(1+x+x^2)(1-x+x^2)=(x^2+1)^2-x^2=1+x^2+x^4$. So: multiply both sides by $1+x+x^2$:

$$\begin{array}{rcl}(1+x+x^2)\cdot(\text{LHS})&\equiv&(1+x+x^2)(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^2+x^4)(1-x^2+x^4)(1-x^4+x^8)\cdots\\&=&(1+x^4+x^8)(1-x^4+x^8)\cdots\\&&\vdots\end{array}$$

The second row shows that the formal product on the left has no (nonzero) odd terms. The third row shows that it doesn't even have the terms $x^2, x^6$ etc. i.e. all the (nonzero) terms are of the form $x^{4k}$, then in the next step all the (nonzero) terms will actually be of the form $x^{8k}$ etc. - so by using induction (or otherwise) you can easily see that the formal product does not have any nonzero terms (except for the constant, which is obviously $1$). Thus, this formal product is $1$.