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I solved one problem in our school math competition. And I think I find the answer in more general case. But I can't prove this. I need to solve the following problem to complete my proof.

  1. Let S=$\{1 < d_1 < d_2 < ... < d_m < n \}$ is the set of all divisors of $n \in \mathbb{N}$. Here $m=\sigma_0(n)-2$. ($\sigma_m(x)$ - Divisor Function. )

  2. Let $k: 1 \leq k < [m/2] $ is an integer and $D_k=d_1 \cdot d_2 \cdot ... \cdot d_k$.

  3. The equation $D_k^4=n^{4k-m}$ can be solved only if m=3, k=1.

The last 3) is my guess. I can't prove this. May be I'm wrong.

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Sorry I don't understand your question, is 3) the question you're trying to solve? –  anegligibleperson Aug 16 '12 at 14:41
    
Yes! I try to solve this equation. My guess is, that it can't be solved in other cases (unless m=3,k=1). –  Mike Aug 16 '12 at 14:45
    
If $n=2^{12}$, then $S=\{ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096\}$ and m=11. So it is not a counterexample. –  Mike Aug 16 '12 at 14:50
    
@Mike Oops, I didn't see that $m$ wasn't free! Let me think some more. –  Cocopuffs Aug 16 '12 at 14:51
1  
@tomasz It holds for $n = $ a prime power, since assuming $n = p^{m+1}$ and $D = p*...*p^k$ one gets a Diophantine equation: $8k^2 + 1$ must be a square. The only possibility for this is $k = 1$, which leads to $m = \frac{1}{2}(\sqrt{8k^2 + 1} + 4k - 1) = 3$. I don't know how to generalize this. –  Cocopuffs Aug 16 '12 at 15:23

2 Answers 2

up vote 5 down vote accepted

My comment above was incorrect; I relied on Wolfram Alpha which gave me a false answer. There are in fact infinitely many solutions, for example $k = 35$ which gives $m = 119$. Indeed, take $n = 2^{120}$, $m = 119$ and $k = 35$. This gives a counterexample to 3, since $$D_k^4 = (2*...*2^{35})^4 = 2^{2*35*36} = 2^{2520} = 2^{120*(4*35 - 119)} = n^{4k - m}$$

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and I believe that this is the smallest counterexample but would be interested if it isn't –  Cocopuffs Aug 16 '12 at 20:48
    
I will leave the comment for reference. Everyone makes mistakes, especially me. –  Cocopuffs Aug 16 '12 at 21:02
    
What a wonderful work! Thanks. I think I will try to find a similar counter-examples based on other primes then 2. My hypothesis was wrong :(. Think I should start another post with my prime problem. –  Mike Aug 16 '12 at 22:49
    
I find the smallest: $n=2^{21}, \ m=20, \ k=6$. $D_6^4=(2\cdot 2^2\cdot....\cdot 2^6)^4=2^{2\cdot 6\cdot 7}=2^{84}$. $n^{4\cdot 6 - 20}=(2^{21})^4=2^{84}$. –  Mike Aug 17 '12 at 1:08
    
@Mike This seems to work, yes –  Cocopuffs Aug 17 '12 at 2:48

Thanks to cocopuffs it is clear now how to construct counterexamples.

Let $n = p^{a+1}$ here p is any prime number. It is easy to see in this case, that $m=a$, $D_k^4=(p \cdot p^2 \cdot ... \cdot p^k)^4=p^{2k(k+1)}$ and $n^{4k-m}=(p^{a+1})^{4k-m}= p^{(m+1)(4k-m)}$. If $D_k^4=n^{4k-m}$ then $(m+1)(4k-m)=2k(k+1)$ and we have Diophantine equation: $2k - m + 4 k m- 2 k^2 - m^2 = 0$ with condition $k<m$ . First solution is $(k=1, m=3)$, second solution is $(k=6, m=20)$.

So $p=2, \ a=m=20, \ k=6, \ n=p^{21}$ is the smallest counterexamples.

By the way $k=35, m=119$ is the 3rd solution.

$$ k_n=\frac{(3 + 2\sqrt{2})^n-(3 - 2\sqrt{2} )^n}{4 \sqrt{2}}, $$

$$ m_n=\frac{1}{4} \left(\left(1+\sqrt{2}\right) \left(3+2 \sqrt{2}\right)^m+\left(1-\sqrt{2}\right)\left(3-2 \sqrt{2}\right)^m -2\right). $$

$k_1=1, \ k_2=6, \ k_3=35, \ k_4=204$ and $m_1=3, \ m_2=20, \ m_3=119, \ m_4=696$.

$k_n=6k_{n-1} - k_{n-2}, \ \ m_n=6m_{n-1} - m_{n-2} + 2$.

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