## Related questions with answers

Question

Show that if k > 1 is an integer, then the equation $\tau(n)=k$ has infinitely many solutions.

Solution

VerifiedStep 1

1 of 2$\hspace*{5mm}$At first for any integer $k>1$, there are infinitely many integers for which

$\tau(n)=k$

$\hspace*{5mm}$Now let

$n=p^{k-1}$

$\hspace*{5mm}$Then if $m$ is any positive integer with prime factorization

$m=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{s}^{a}$

$\hspace*{5mm}$Number of divisors of $m$ is

$\tau(m)=\prod_{j=1}^{s}\left(a_{j}+1\right)$

$\hspace*{5mm}$ $\tau(n)=k$ since there are infinitely many primes therefore there are infinitely many $n$ such that

$n=p^{k-1} \text{ and } \tau(n)=k$

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