Let $$a, b, c, d$$ be positive integers such that $$a^2 + b^2 - cd^2 = 2026$$. Find the minimum possible value of $$a + b + c + d$$.

Since $$cd^2 \ge 1$$ we need $$a^2 + b^2 \ge 2027$$, and for a fixed sum $$a + b \le 46$$ the largest possible value of $$a^2 + b^2$$ is $$45^2 + 1 = 2026$$, so $$a + b \ge 47$$. If the total were at most 50 then $$c + d \le 3$$; with $$c = d = 1$$ we would need $$a^2 + b^2 = 2027 \equiv 3 \pmod 4$$, which is impossible, and with $$c + d = 3$$ the value $$a^2 + b^2$$ is even, forcing $$a + b$$ even and contradicting $$a + b = 47$$. Hence the minimum is 51, attained by $$(a, b, c, d) = (45, 2, 3, 1)$$ because $$45^2 + 2^2 - 3 \times 1^2 = 2026$$.