10.1. (12 points) For different natural numbers $k, l, m, n$, it is known that there exist such three natural numbers $a, b, c$ that each of the numbers $k, l, m, n$ is a root of either the equation $a x^{2}-b x+c=0$, or the equation $c x^{2}-16 b x+256 a=0$. Find $k^{2}+l^{2}+m^{2}+n^{2}$.
Answer: 325.

Solution. If $k, l$ are the roots of the first equation, then the roots of the second equation are the numbers $m=\frac{16}{k}$, $n=\frac{16}{l}$. Therefore, the numbers $k, l, m, n$ are divisors of the number 16. The divisors of 16 are the numbers 1, 2, 4, 8, and 16, but the number 4 does not fit, as by the condition all numbers $k, l, m, n$ are distinct. Thus, $k^{2}+l^{2}+m^{2}+n^{2}=1^{2}+2^{2}+8^{2}+16^{2}=325$.