18. Let non-zero natural numbers $a, b, c$ satisfy the equation $a^{2}+b^{2}-c=2022$. Try to find:
(1) The minimum value of $a^{2}+b^{2}+c^{2}$;
(2) The values of $a, b, c$ when $a^{2}+b^{2}+c^{2}$ takes the minimum value.
【Answer】(1) 2034 (2) $27,36,3$
【Solution】
Since a perfect square divided by 4 can only leave a remainder of 0 or 1, the sum of two perfect squares divided by 4 can only leave a remainder of 0, 1, or 2. From the problem, we have $a^{2}+b^{2}=2022+c$. (Method 1) When $c=2$, $a^{2}+b^{2}=2024$, then $\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}=506$, so $\frac{a}{2} \div 4 \cdots 1, \frac{b}{2} \div 4 \cdots 1$, trying $1,5,9,13,17$, we find no solution.
(Method 2) When $c=2$, $a^{2}+b^{2}=2024$, since $11 \mid 2024$, and a perfect square divided by 11 can only leave a remainder of 0, 1, 4, 9, 5, 3, so $11|a, 11| b$, but 2024 is not a multiple of 121, so there is no solution.
Similarly, using Method 2, we can conclude that $a, b$ are both multiples of 9.
When $c=3$, $a^{2}+b^{2}=2025=45^{2}=3^{2} \times 5^{2}=\left(3^{2}+4^{2}\right) \times 3^{-1}=27^{2}+36^{2}$. At this point,
$a^{2}+b^{2}+c^{2}=2025+9=2034, a=27, b=36, c=3$ or $a=36, b=27, c=3$