November 15th, 2017, 01:55 PM  #1 
Newbie Joined: Nov 2017 From: Mexico Posts: 1 Thanks: 0  Diophantine hyperbola
I am given an integer c an and we know the sum of consecutives integers can reach this number. I reach this expression \begin{align*} c &= \dfrac{n(n+1)}{2}  \dfrac{m(m+1)}{2} \end{align*} where n > m, both are integers positive. we can reduce that to \begin{align*} 2 * c &= n^2 + n  m^2  m \end{align*} For example if c = 1000 then one of the solutions is n = 52 and m = 27. How can I get all the integer solutions n and m ? Last edited by limboa; November 15th, 2017 at 02:04 PM. Reason: v8archie was right 
November 15th, 2017, 02:01 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,142 Thanks: 2382 Math Focus: Mainly analysis and algebra 
${}m$, not ${}+m$

November 15th, 2017, 02:55 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,142 Thanks: 2382 Math Focus: Mainly analysis and algebra 
Not sure it helps, but \begin{align*}n^2+nm^2m&=\frac14\left((2n+1)^2  (2m+1)^2\right) \\ &=\frac14(2n+2m+2)(2n2m)=(n+m+1)(nm) \end{align*}


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diophantine, hyperbola 
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