Q:

Does a polynomial solution for an NP-complete problem that can only be implemented for small N *still* imply P=NP?

Basic sanity check on NP-complete solutions:
Suppose there was a polynomial time solution for an NP-complete problem,
but the size of NP-complete problems that could be solved is still
relatively small (i.e. N = 32-64) due to limits in technology.
Would this show that P = NP? Or would this be more like the
pseudo-solution that the unary subset sum is in P?

A:

If you mean that the polynomial-time algorithm only works for inputs up to some fixed size, it shows nothing at all. Any problem at all (even if it's undecidable, let alone NP-complete) becomes a finite language when restricted to instances of constant size. All finite languages can be decided in constant time.
If you mean that the polynomial-time algorithm works for all inputs but it's still so inefficient that current computers can only run it on small instances then, sure, that would prove that P$\,=\,$NP. The idea that "polynomial" means "efficiently and effectively solvable" is just a short-hand. The actual definitions of P and NP don't depend on that short-hand.

A:

Time complexity "for small inputs" simply doesn't make sense, because the definition of time complexity is based on the limit of the running time as the input size grows to infinity.

A:

I present one possible interpretation of your question such that the answer is "YES."
Suppose that you could solve the satisfiability problem in polynomial time, but the leading term of the polynomial had a large exponent, e.g. $n^{20}$. Then it would be, indeed, the case that the solution is not practical, i.e., you can only solve the satisfiability problem on small instances using this hypothetical algorithm. Nonetheless, NP would still collapse to P. Sometimes, this exact reasoning is used to point out that the definition of P does not quite capture the notion of "practical algorithms," to which the response is that algorithms that we come up with tend to have small exponents of leading terms.

