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Question: Find all the couples of $(m,n)\in \mathbb{Z^+ \times Z^+}: \tau(m)=10,\ \tau(n)=21$ and $\gcd(m,n)=18.$

Answer: Let $m=p_1^{m_1}\cdots p_k^{m_k},\ n=q_1^{n_1}\cdots q_l ^{n_l}$ be the prime factorization of $m,n$.

  • $\tau(m)=10\iff (m_1+1)\cdots(m_k+1)=10\implies m_i+1|10,\ \forall i=1,...,k \implies m_i+1=1,2,5,10\implies m_i= 0,1,4,9,\ \forall i=1,...,k $.

  • $\tau(n)=21 \iff... \iff n_j=0,2,6,20, \forall j=1,...,l $

From these, we can see that the forms of $m,n$ are $n=p_1^2p_2^6$ or $p^{20}$ and $m=q_1q_2^4$ or $q^9$. And the only case to take $\gcd(m,n)=18=2\cdot3^2$ is if $m=2 \cdot 3^4 $ and $n=3^2\cdot 2^6$.

Is this the correct answer?

Thank you.

  • 1
    The answer is correct. The calculations are not too accurate though. $10=2\cdot 5$ for $\tau(m)$ and $21=3\cdot 7$ for $\tau(n)$, how come you conclude $m=p_1^{2}p_2^{6}$ (should be $p_1^{1}p_2^{4}$) and $n=q_1^1q_2^4$ (should be $n=q_1^2q_2^6$)? – 2017-02-06
  • 0
    @rtybase Thank you for your comment. I had a typo (m & n were in opposite places). Now, is it clearer? – 2017-02-06
  • 1
    Yep, much better ;) – 2017-02-06

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