Prove that $f(n)=n^2$ where $f$ is a strictly increasing multiplicative function with $f(2)=4$.

Question

Let $f:\mathbb N\to\mathbb N$ be a strictly increasing function with $f(2)=4$ which is completely multiplicative i.e $f(ab)=f(a)f(b)$ for all $a,b\in\mathbb N$. Prove that $f(n)=n^2$ for all $n\in\mathbb N$.

This is an exercise on induction. So I am looking for an inductive solution. Here is my progress:

It is easy to see that $f(1)=1$. For the base case, we need to find $f(3)$ which I had some difficulties to find. Now $$ f(3^2) > f(2^3) = 64 \implies f(3)>8\\ f(3^8)<f(2^{13})=67108864\implies f(3)<10 $$

So $8<f(3)<10$ or $f(3)=9$. Now I can show that $f(2^k)=4^k$ for all $k\in\mathbb N$. Then I tried to prove $f(n+1)=(n+1)^2$ assuming $f(i)=i^2$ for all $i\leq n$. But this doesn't work.

So how do I solve the problem?

Answer 1

You have already shown the cases $n=1,2$. Now we want to do the induction step. To simplify notation I write $m=n+1$. If $m$ is not prime, we are done (as we can just use the induction hypothesis on the factors). Thus, in the following we will assume that $m$ is prime. First we show a lower bound

$$ f(m)^2=f(m^2) > f(m^2-1)=f(m+1) f(m-1) = (m+1)^2 (m-1)^2 = (m^2-1)^2.$$ We used that $m$ is prime and thus $m\pm 1$ are divisible by $2$, which implies that we can use the induction hypothesis to show $f(m\pm 1)=(m\pm 1)^2$. Taking square root yields $f(m)>m^2-1$.

Next we show the upper bound. Let $N\in \mathbb{N}_{\geq 1}$, then there exists a unique $\ell(N)\in \mathbb{N}$ such that $$ 2^{\ell(N)} \leq m^N < 2^{\ell(N)+1}.$$ Then we have $$ m^N < 2^{\ell(N)+1} \leq 2 m^N. $$ Thus, we obtain $$ f(m)^N < f(2^{\ell(N)+1}) = (2^{\ell(N)+1})^2 < 4 m^{2N}. $$ After taking the $N$th root, we get $$ f(m) \leq 4^{1/N} m^2. $$ Hence, for $N\rightarrow \infty$ we get $$ f(m) \leq m^2. $$ Combining the upper and the lower bound we get $f(m)=m^2$.

Answer 2

Disclaimer: Not quite sure if this is the inductive proof that the exercise is going for, since that's not the crux step.

1. Prove by induction that $f(2^k) = 2^{2k}$ for $ n \geq 0$.

2. Suppose that $ f(n) \geq n^2 + 1$ for $ n \neq 2^i$, then

Find a $k, m \in \mathbb{N}$ such that $(n^2)^m < 2^k < (n^2+1)^m$.
Why must such a $k,m$ exist?

All that we need is $ \log n^2 < \frac{k}{m} \log 2 < \log (n^2 + 1) $, and clearly we can find such a fraction $ \frac{k}{m}$ to satisfy the inequality.

Hence, reach the contradiction:

$$ (n^2+1)^{2m} \leq f(n)^{2m} = f(n^{2m} ) < f(2^k) = 2^{2k} < (n^2+1)^{2m}.$$

So, $ f(n) \leq n^2$.

Note: This is essentially what you did with $9^ 4 < 2^{13} < 10^4$ to conclude that $f(3) < 10$.

3. Likewise for $ f(n) \leq n^2 -1 $ we have a contradiction.

Note: This is essentially what you did with $ 8 \leq 2^3 < 9$ to conclude that $8 < f(3) $.
(Just that I'm enforcing strict inequality on both sides. Obviously inequality on one side is enough and will always happen, but I didn't want to complicate with a minor detail.)

4. Hence $f(n) = n^2$.

---

This is much uglier, and still not very induction-y

1. Prove by induction that $f(2^k) = 2^{2k}$ for $ n \geq 0$.

2. Prove by strong induction on $n \geq 4$ that $f(n) = n^2$.
   If $n=ab$ is not a prime, then we have $f(n) = f(ab) = f(a)f(b) = a^2b^2 = n^2 $.
   If $ n \geq 4$ is a prime, then $ n-1, n+1$ have prime factors $<n$ so $ f(n-1) = (n-1)^2, f(n+1) = (n+1)^2$. Thus $(n-1)^2 < f(n) < (n+1)^2$.

We now consider how to bound $ f( n^{2^k} ) = f(n) ^{2^k} $ using $ f(n+1), f(n-1)$ to tighten the inequality further.

For all $k$, there is a unique integer $a_k $ such that
$$ (n+1)^ {a_k} (n-1) ^ { 2^k - a_k } < n^{2^k} < (n+1)^{a_k +1} (n-1) ^ {2^k - a_k - 1 }, $$

or that

$$ \frac{ a_k \log (n+1) + (2^k - a_k ) \log(n-1) } { 2^k } < \log n < \frac{ (a_k+1) \log (n+1) + (2^k - a_k -1 ) \log(n-1) } { 2^k }$$

Notice that the difference between the extreme ends of the inequality is $ \frac {\log(n+1) - \log(n-1) } { 2^k}$ which tends to 0, the LHS is bounded above by $\log n$, the RHS is bounded below by $ \log n$, hence both of their limits are $ \log n$.

This gives us

$$ (n+1)^{2a_k} (n-1)^{2(2^k-a_k) } < f(n)^{2^k} < (n+1)^{2(a_k+1)} (n-1)^{2(2^k-a_k-1) },$$

or that

$$ \frac{ (2a_k) \log(n+1) + 2(2^k- a_k ) \log(n-1)}{2^k} < \log f(n) < \frac{(2a_k) \log(n+1) + 2(2^k- a_k ) \log(n-1)}{2^k} $$

Since the LHS and RHS have the same limit of $ 2 \log n$, it follows that $ \log f(n) = 2 \log n \Rightarrow f(n) = n^2$.