Let f: N → N be a function such that f(m + n) = f(m) + f(n) for...

Given functional equation is

$f\left(m + n\right) = f\left(m\right) + f\left(n\right); m, n \in N . . . . \left(1\right)$

Put $m = n = 3$

$f ( 3 + 3 ) = f ( 3 ) + f ( 3 )$

$⇒ f ( 3 ) = 9\left[∵ f\left(6\right) = 18\right]$

Put $m = 2, n = 1\text{in equation}\left(1\right)$

$∴ f ( 3 ) = f ( 2 + 1 ) = f ( 2 ) + f ( 1 )$

$= f ( 1 + 1 ) + f ( 1 )$

$= f ( 1 ) + f ( 1 ) + f ( 1 )$

$9 = 3f ( 1 )$

$⇒ f ( 1 ) = 3$

$∴ f ( 2 ) = f ( 1 + 1 ) = f ( 1 ) + f ( 1 ) = 6$

Now,

$f ( 2 ) \cdot f ( 3 ) = ( 6 ) ( 9 ) = 54$