Modular quadratic equation (solve for 3-digit natural numbers)

Question

$n^2 + 6n - 88$ is divisible by 97. Solve for all n if n is a 3-digit natural number.

Here's my progress so far

$$n^2 + 6n - 88 \equiv 0\pmod {97}$$ $$n^2 + 6n - 88 + 97 \equiv 0\pmod {97}$$ $$n^2 + 6n + 9 \equiv 0\pmod {97}$$ $$(n+3)^2 \equiv 0\pmod {97}$$

I know $97$ is prime, but can't figure out how to use this fact. I can't just take the sqrt, can I?

Any hints, tips or anything else highly appreciated.

Answer 1

You are almost there. Your last step implies that (n+3) is divisible by 97 (since if a² is divisible by b, a should also be divisible by b. Just list all three digit factors of 97, and subtract 3 from each to get to the answer. So your answer is 191, 288, 385, 482, 579, 676, 773, 870, 967.