combinatorics - $sum_{k=0}^cbinom{a-k(p-1)}{k}binom{b+k(p-1)}{c-k}=sum_{k=0}^c(-p)^kbinom{a+b+1}{c-k}$ is true?

Question : Let $a,b,c$ be non-negative integers. Also, let $p$ be a prime number. Then, is the following true?
$$\sum_{k=0}^c\binom{a-k(p-1)}{k}\binom{b+k(p-1)}{c-k}=\sum_{k=0}^c(-p)^k\binom{a+b+1}{c-k}.$$

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Motivation : I found the following question in a book :
(The conditions about $a,b,c,p$ are the same as above.)

Prove the following in mod $p$:
$$\sum_{k=0}^c\binom{a-k(p-1)}{k}\binom{b+k(p-1)}{c-k}\equiv\binom{a+b+1}{c}$$

This question can be solved by either induction or using
$$\sum_{k=0}^c\binom{a-k(p-1)}{k}\binom{b+k(p-1)}{c-k}\equiv\sum_{k=0}^c\binom{a+k}{k}\binom{b-k}{c-k}.$$
(Also, we know that $p$ does not need to be a prime number in this question.)

After thinking about this question, I reached the above expectation. Can anyone help?

Answer

The equality above is between polynomials in $p$, of degree $c$. If it is true for many primes it is true for all numbers $p$. But for $p=0$, $c=1$ we get:

$$\binom{a}{0}\binom{b}{1}+\binom{a}{1}\binom{b}{0}=a+b\neq a+b+1=\binom{a+b+1}{1}$$