# Question on generating function of integer partitions

How can I show that $$\prod_{k \ge 1}(1+z^{2k}) = \prod_{k \ge 1}(1+z^k+z^{2k}+z^{3k}) \quad ?$$

I have worked on this for a while and I am even doubting that maybe both are not equal

• Compare the coefficients of $z$, or for that matter any $z^{2n+1}$, on both sides... – Sam Streeter Oct 24 '18 at 20:10
• If we divide both sides by $\prod_{k\geq 1}(1+z^{2k})$ we get $$1 = \prod_{k\geq 1}(1+z^k)$$ which is certainly not true. – Jack D'Aurizio Oct 24 '18 at 20:14
• @JackD'Aurizio . I got The LHS as the generating function for the set of partions of $n$ which every even part appear at most once and the RHS as the generating function for the set of partitions of $n$ in which every part appears at most three times. Did I interpret this wrong? My goal is to show that both are equal. – Jaynot Oct 24 '18 at 20:18
The generating function for the partitions in which each part appears at most three times is $$\prod_{k\geq 1}(1+x^k+x^{2k}+x^{3k})$$, correct, but the generating function for the partitions in which every even part appears at most once is $$\prod_{k\geq 1}(1+x^{2k})\prod_{k\geq 1}\frac{1}{1-x^{2k-1}}$$ since the odd parts may appear as many times as they want. Then we have to check that
$$\prod_{k\geq 1}\frac{1+x^{2k}}{1-x^{2k-1}}=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k}$$ which is equivalent to $$\prod_{k\geq 1}\frac{1-x^k}{1-x^{2k-1}}=\prod_{k\geq 1}\frac{1-x^{4k}}{1+x^{2k}}$$ or to $$\prod_{k\geq 1}\frac{1-x^k}{1-x^{2k-1}}=\prod_{k\geq 1}(1-x^{2k})$$ which is trivial since $$\prod_{k\geq 1}(1-x^k) = \prod_{k\geq 1}(1-x^{2k})(1-x^{2k-1})$$ (the $$k$$ on the left is either odd or even).