There are infinitely many ordered pairs
(m,n) of positive integers for which
m + (m+1) + (m+2) + ... (n-1) + n = mn.
List the first five pairs ordered by values of m.
I'll start with Charlie's equation n^2 - m^2 + m + n - 2nm = 0.
Rearrange this into (m+n)^2 - 1*(m+n) +(1/4)n^2= (9/4)*n^2.
Then m+n-(1/2) = sqrt[2n^2+(1/4)]
Eliminating fractions and roots yields (2m+2n-1) - 8n^2 = 1
This Pell equation has a recursive solution with
y=n is simple, so that leaves m. 2m+2n-1 = x and y=n then implies m = (x+1)/2-y. Using induction m_t = 6*m_(t-1) - m_(t-2) - 2
OEIS A011900
So, the sequence of ordered pairs begins with (1,1), (3,6), (15,35), (85,204), (493,1189), ....
Solution
