We know that the Chi Square PDF is $(x^{-1 + n/2}\cdot 2^{-n/2}\cdot e^{-x/2})/\Gamma[n/2]$
We know that Gamma PDF is $(\frac{x}{b})^{-1 + a}\cdot e^{-(x/b)}/(b\cdot\Gamma[a])$
Given an n, how can we set a and b such that the Gamma PDF is the same as the Chi Square PDF? My first thought is that $a = n/2$, but I am having trouble from there.
Good thought.
When $a=n/2$ then $$(\frac{x}{b})^{-1 + a}\cdot e^{-x/b}/(b\cdot\Gamma[a])\\ = x^{-1 + n/2}\cdot b^{-n/2}\cdot e^{-x/b}/\Gamma[n/2]$$
You want this to equal: $$x^{-1 + n/2}\cdot \color{red}{2}^{-n/2}\cdot e^{-x/\color{red}{2}}/\Gamma[n/2]$$
Why are you having trouble?