Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.

Solution:

Let the larger and smaller parts be $x$ and $y$, respectively.

According to the question:

$x+y=16 \ldots$ (i)

$2 x^{2}=y^{2}+164 \ldots$ (ii)

From (i), we get:

$x=16-y \ldots$ (iii)

From (ii) and (iii), we get:

$2(16-y)^{2}=y^{2}+164$

$\Rightarrow 2\left(256-32 y+y^{2}\right)=y^{2}+164$

$\Rightarrow 512-64 y+2 y^{2}=y^{2}+164$

$\Rightarrow y^{2}-64 y+348=0$

$\Rightarrow y^{2}-(58+6) y+348=0$

$\Rightarrow y^{2}-58 y-6 y+348=0$

$\Rightarrow y(y-58)-6(y-58)=0$

$\Rightarrow(y-58)(y-6)=0$

$\Rightarrow y-58=0$ or $y-6=0$

$\Rightarrow y=6 \quad(\because y<16)$

Putting the value of y in equation (iii), we get:

$x=16-6=10$

Hence, the two natural numbers are 6 and $10 .$