To give context this is Problem 1.1.8 from Putnam and Beyond
I have a simple solution for this problem that I would like to share. I checked the book's solution, and although it is very clever, it is also very convoluted (involving the density of rationals on the real line at some point). I would appreciate any feedback and I also would love see your solution to this problem.
Show that there does not exist a strictly increasing function $f : \mathbb{N} \rightarrow \mathbb{N}$ satisfying $f(2) = 3$ and $f (mn) = f(m)f(n)$ for all $m, n ∈ N$.
My solution to this problem:
Let $f$ be such a function, then $f(9) = f(3)^2>f(8)=27$ and so $f(3)>5$. And if $f(3) \ge 7$ then $f(10) > f(9) \ge 49$ which in turn implies that $f(5) \ge \dfrac{f(10)}{f(2)} \ge 17$. Now we can show that $f(15) =f(3)f(5)\ge 7\cdot 17 \ge 119 > f(16)= 81$, and so $f(3) \lt 7$. Therefore, $f(3)=6$.
Here comes the difficult part of the problem. First notice that $f(7) = \dfrac{f(28)}{f(4)} > \dfrac{f(27)}{f(4)} =24$ and so $f(7) \ge 25$ and $f(5)>\dfrac{f(9)}{f(2)}=12$ and so $f(5) \ge 13$. From this we can conclude that, $f(35) =f(7)f(5)\ge 25\cdot 13 \ge 325 > f(36)=324$. Contradiction!
Although this is very elementary proof, it took me $1.5$ hours of scratch-work.