How to solve this?

  • Thread starter Sam_
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  • #26
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(m+1)/n+(n+1)/m=3 also has an infinite number of solutions. I believe that is it.
 
  • #27
Ben Niehoff
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I have it. Suppose

[tex]\frac{m+1}n + \frac{n+1}m = Z[/tex]

where Z is some integer. Then

[tex]m^2 + m + n(n+1) = Zmn[/tex]

[tex]m^2 - (Zn - 1)m + n(n+1) = 0[/tex]

This has two roots

[tex]m^2 - (m_1 + m_2)m + m_1m_2 = 0[/tex]

Therefore, we have

[tex]m_1 + m_2 = Zn - 1[/tex]

But we know that

[tex]Z = \frac{m_1+1}n + \frac{n+1}{m_1}[/tex]

[tex]Zn = m_1 + 1 + \frac{n(n+1)}{m_1}[/tex]

and so

[tex]m_1 + m_2 = Zn - 1 = m_1 + \frac{n(n+1)}{m_1}[/tex]

which yields

[tex]m_2 = \frac{n(n+1)}{m_1}[/tex]

Now we just have to prove this is an integer. First, we know:

[tex]\frac{m_1+1}n + \frac{n+1}{m_1} = Z[/tex]

[tex]m_1(m_1+1) + n(n+1) = Zm_1n[/tex]

[tex]n(n+1) = Zm_1n - m_1(m_1+1)[/tex]

[tex]\frac{n(n+1)}{m_1} = Z - m_1 - 1[/tex]

which is an integer. QED

Edit: This doesn't technically prove that there are infinitely many solutions, because I haven't proved that m_2 is always greater than m_1...hmm...
 
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  • #28
Ben Niehoff
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Suppose Z is a positive integer (if m and n are positive, this is always the case). Then suppose that m_1 < n. Then if Z > 2,

[tex]Zn - m_1 - 1 > Zn - n - 1 = (Z-1)n - 1 > n > m_1[/tex]

Now all we need is at least one case for Z > 2 and m < n, which Sennyk has already shown.

QED
 
  • #29
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Relationship between solutions

I've found the recursive relationship between the solutions.

case [tex]\frac{m+1}{n}+\frac{n+1}{m}=4[/tex]

[tex]n_0=1, m_0=1[/tex]
[tex]n_1=1, m_1=2[/tex]
[tex]n_k=m_k_-_1, m_k=\frac{n_k(n_k+1)}{m_k_-_2}[/tex]

case [tex]\frac{m+1}{n}+\frac{n+1}{m}=3[/tex]

[tex]n_0=2, m_0=2[/tex]
[tex]n_1=2, m_1=3[/tex]
[tex]n_k=m_k_-_1, m_k=\frac{n_k(n_k+1)}{m_k_-_2}[/tex]

Maybe a proof by induction would work?
 
  • #30
Ben Niehoff
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Yeah, my above two posts give a proof by induction. That recursive relation shows up in the proof.

Part of the proof involves proving that

[tex]m_k = \frac{n_{k-1}(n_{k-1} + 1)}{m_{k-2}}[/tex]

is always an integer. Finally, one also has to prove that

[tex]m_k > m_{k-2}[/tex]

or else the attempt at "induction" might actually loop back on itself.
 
  • #31
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I have it. Suppose

[tex]m^2 - (Zn - 1)m + n(n+1) = 0[/tex]

This has two roots

[tex]m^2 - (m_1 + m_2) + m_1m_2 = 0[/tex]
I'm not seeing how you make this leap. Please explain.
 
  • #32
Ben Niehoff
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I'm not seeing how you make this leap. Please explain.
Whoops, there's a typo. It should be:

[tex]m^2 - (m_1 + m_2)m + m_1m_2 = 0[/tex]

It's directly from the fundamental theorem of algebra. If a second-degree polynomial has roots r_1 and r_2, then

[tex]\begin{array}{rcl}(x - r_1)(x - r_2) & = & 0 \\ x^2 - r_1x - r_2x + r_1r_2 &=& 0 \\ x^2 - (r_1 + r_2)x + r_1r_2 &=& 0\end{array}[/tex]
 
  • #33
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sennyk, while he did make a typo, you have got to look at symmetric functions as they apply to the roots of a polynominal.

Take the equation X^3-1 = 0. This equation has three roots x=1, [tex]X=\frac{-1\pm\sqrt-3}{2}[/tex]

Question: What is the sum of the three roots and what is their product?

Answer: In the equation X^3-bX^2+cX-d, the sum of the roots equals -b, and the product of the roots equals -d. Since b=0 the sum of the roots is 0 and since -d =1 the product of the roots is 1.

We get the above form from multiplying out (x-r)(x-s)(x-t), where r,s,t represent the three roots.
 
Last edited:
  • #34
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The typo completely threw me off. I know how to find the roots of a polynomial. I'm not a complete amateur. :)
 
  • #35
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O.K., sorry
 

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