Are there identities on integers that be proved by checking finitely many cases?

65 Views Asked by At

Some time ago, I needed to prove some identity on binomial coefficients. Something similar to the following:

$$\sum_{m=0}^n \binom m j \binom {n-m}{k-j}= \binom {n+1}{k+1}$$

but with somewhat different coefficients. I could not find a formal proof, but I computed the values for $j,k,n$ up to 1000 and found that it holds.

Now, I know that formally this is not sufficient for a proof, since the the identity might fail on 1,001. However, the identity looks so "simple" (in that it uses only multiplications and additions - no roots, prime numbers etc.), that it seems impossible that it would start failing after 1,000.

Is there a way to make this intuition formal? I.e., is there a class of identities on integers that are considered so simple, that proving an identity from that class for sufficiently many integers is sufficient for proving that it holds for all integers?

Alternatively, is there a binomial identity of a similar simple form, which is known to hold up to some very large integer, but fails afterwards?