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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 $2^k-1=v_p(f(2^n))<v_p(f(2^{n+1}))=2^{\ell}-1$. 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?

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

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  • $\begingroup$ So it is true for any prime, not just 2? $\endgroup$ – user180446 Jun 27 '20 at 19:34

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