# The smallest integers having $2^n$ divisors

Problem: For any integer $$d > 0,$$ let $$f(d)$$ be the smallest possible integer that has exactly $$d$$ positive divisors (so for example we have $$f(1)=1, f(5)=16,$$ and $$f(6)=12$$). Prove that for every integer $$k \geq 0$$ the number $$f\left(2^k\right)$$ divides $$f\left(2^{k+1}\right).$$

My Solution: We begin by observing that there must exist a prime $$p$$ such that $$2^k-1=v_p(f(2^n)). Otherwise, we have $$v_p(f(2^n))\ge v_p(f(2^{n+1}))$$ for all prime. But this isn't possible, since $$f(2^{n+1})$$ has more divisors than $$f(2^{n})$$ (by definition). Now consider the number $$N=\frac{f(2^{n+1})}{p^{2^{\ell-1}}}$$. This number has $$2^n$$ divisors. So we must have $$N\ge f(2^n)$$. Now consider the number $$f(2^n)p^{2^k}$$. This number has $$2^{n+1}$$ divisors. So we must $$f(2^n)p^{2^k}\ge N\cdot p^{2^{\ell-1}}\ge f(2^n)p^{2^{\ell-1}}$$. Thus we must have $$N=f(2^n)$$ and $$\ell=k+1$$. Thus we must have $$f(2^n)\mid f(2^{n+1})$$.

But If my solution is correct, then we have nothing special about 2. Thus I am skeptic wheteher my proof is correct or not. Can someone please point out any error?

• At the risk of sounding stupid, what is $v_p$? – orlp Jun 27 '20 at 18:49
• Sorry for not clarifying this non-standard notation. Suppose $p^a\mid n$ and $p^{a+1}\nmid n$, then $v_p(n)=a$. – user180446 Jun 27 '20 at 18:51
• @orlp See $p$-adic order. – John Omielan Jun 27 '20 at 18:55
• This is fairly standard notation for the $p$-adic order or valuation, except that it should be $\nu_p$ (Greek letter nu). – TonyK Jun 27 '20 at 18:56
• I think that $f(p^k)$ is simply the product of the $(p-1)$th powers of the first $k$ primes. So $f(p^k)$ divides $f(p^{k+1})$ for all primes $p$. – TonyK Jun 27 '20 at 18:58

You do use something special about $$2$$, namely that it is prime:
If you replace $$2$$ with a different prime $$q$$, you end up with $$f(q^n)p^{q^{k+1}-q^k}\ge f(q^{n+1})=Np^{q^{\ell}-q^{\ell-1}}\ge f(q^n)p^{q^{\ell}-q^{\ell-1}},$$ so $$k\ge \ell-1$$ and again $$k=\ell-1$$ and $$f(q^{n+1})=p^{(q-1)q^k}f(q^n)$$.