I am trying to show that$$(1 + x + x^2 +x ^3)^p = \sum_{n \ge 0} \sum_{m=0}^{n} {p \choose m}{p \choose n - 2m}x^n$$
I tried $(1 + x + x^2 + x^3)^p = (\frac{1-x^4}{1-x})^p = (1-x^4)^p (1-x)^{-p}$ following $$(1-x^4)^p = \sum_{j=0}^{\infty} (-1)^j {p \choose j}x^{4j} \quad \text{and} \quad (1-x)^{-p} = \sum_{i=0}^{\infty} {p+i-1 \choose p-1} x^i$$ I am not sure if thats the correct way so any help is appreciated
On
We can also start from the right-hand side and transform it to obtain the left-hand side. Since $n$ occurs in only one binomial coefficient, a starter is to exchange the sums and try to continue this way.
We obtain \begin{align*} \color{blue}{\sum_{n=0}^{\infty}}&\color{blue}{ \sum_{m=0}^{n} \binom{p}{m}\binom{p}{n-2m}x^n} =\sum_{0\leq m\leq n<\infty} \binom{p}{m}\binom{p}{n-2m}\tag{1}\\ &=\sum_{m=0}^\infty\sum_{n=m}^{\infty}\binom{p}{m}\binom{p}{n-2m}x^n\tag{2}\\ &=\sum_{m=0}^\infty\binom{p}{m}\sum_{n=2m}^{\infty}\binom{p}{n-2m}x^n\tag{3}\\ &=\sum_{m=0}^\infty\binom{p}{m}\sum_{n=0}^{\infty}\binom{p}{n}x^{n+2m}\tag{4}\\ &=\sum_{m=0}^\infty\binom{p}{m}\left(x^2\right)^m\sum_{n=0}^{\infty}\binom{p}{n}x^{n}\tag{5}\\ &=\left(1+x^2\right)^p(1+x)^p\tag{6}\\ &\,\,\color{blue}{=\left(1+x+x^2+x^3\right)^p} \end{align*} and the claim follows.
Comment:
In (1) we use a convenient notation to better see the change of summation in the following line.
In (2) we exchange the summation order.
In (3) we factor out $\binom{p}{m}$ and start the inner sum with $n=2m$ noting that $\binom{\alpha}{k}=0$ if $k < 0$.
In (4) we shift the index of the inner sum to start with $n=0$.
In (5) we factor out $x^{2m}$.
In (6) we observe we can apply the binomial theorem twice noting the upper index of both sums is $p$ since other terms with $m,n>p$ vanish.
Hint: $$1+x+x^2+x^3=(1+x)(1+x^2)$$