<s>[INST] Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$. 
A000332 has the comment:

Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.

The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$): 
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
 [/INST] The identity is equivalent to
\begin{align*}
n&=\binom{3}{0}\binom{n + 3}{n-1}-\binom{3}{1} \binom{n + 2}{n-2} + \binom{3}{2}\binom{n + 1}{n-3} - \binom{3}{3}\binom{n}{n-4}\\
&=\sum_{k=0}^3 (-1)^k\binom{3}{k}\binom{(n-k)+3}{(n-k)-1}=[x^n](1-x)^3\cdot \frac{x}{(1-x)^5}=[x^{n+1}]\frac{1}{(1-x)^2}
\end{align*}
which holds.
P.S. Note that by replacing $3$ with $m$, we get
$$\sum_{k=0}^m (-1)^k\binom{m}{k}\binom{(n-k)+m}{(n-k)-1}=[x^n](1-x)^m\cdot \frac{x}{(1-x)^{m+2}}=[x^{n+1}]\frac{1}{(1-x)^2}=n.$$
</s>