Find all possible values of natural numbers $$m$$ and $$n$$ such that $$12mn - 175 = 105m + 140n$$.

Rewrite the equation as $$12mn - 105m - 140n + 1225 = 1400$$, which factorises as $$(3m - 35)(4n - 35) = 1400 = 2^3 \times 5^2 \times 7$$. Both factors cannot be negative, because $$3m - 35 \ge -32$$ and $$4n - 35 \ge -31$$ would make the product at most 992. So both are positive, with $$4n - 35 \equiv 1 \pmod{4}$$ (hence odd) and $$3m - 35 \equiv 1 \pmod{3}$$. The odd divisors of 1400 are 1, 5, 7, 25, 35, 175, and those that are $$1 \pmod{4}$$ are 1, 5, 25, giving $$3m - 35 = 1400, 280, 56$$ respectively, of which only 280 is $$1 \pmod{3}$$. Thus $$4n - 35 = 5$$ and $$3m - 35 = 280$$, so the only solution is $$m = 105$$, $$n = 10$$ (check $$12600 - 175 = 12425 = 11025 + 1400$$).