The n−th triangular number is given by $\displaystyle T_{n} = \frac{1}{2}n(n+1)$. how can one find all integers n for which the sum of the (n − 1)−st, the n−th,
and the (n + 1)−st triangular numbers is a square?
Google told me to look in the papers:
Triangular Numbers and Perfect Squares
Tom Beldon and Tony Gardiner
The Mathematical Gazette
Vol. 86, No. 507 (Nov., 2002), pp. 423-431
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88.53 Comments on 'Triangular Numbers and Perfect Squares'
Michael D. Hirschhorn
The Mathematical Gazette
Year, volume, issue, and the page range: Vol. 88, No. 513 (Nov., 2004), pp. 500-503
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We have $\displaystyle 3n^2+3n+2=2m^2 $
Now multiply by $\displaystyle 12 $ and add and subtract 9 to get $\displaystyle (6n+3)^2-24m^2+15=0 $
Let $\displaystyle y=6n+3 $ and multiply both sides by $\displaystyle 24 $ to get $\displaystyle (24m)^2-24y^2=360 $
By letting $\displaystyle x=24m $ we see we have $\displaystyle x^2-24y^2=360 $ which is almost Pell's Equation.
Luckily though this is solved similarly. I won't solve it, but here for reference on how to solve it (look for the part about solving $\displaystyle x^2-ny^2=c $).