# Thread: Working with triangular numbers

1. ## Working with triangular numbers

The n−th triangular number is given by $T_{n} = \frac{1}{2}n(n+1)$. 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?

2. Originally Posted by Intsecxtanx
The n−th triangular number is given by $T_{n} = \frac{1}{2}n(n+1)$. 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

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

In case you do not have access, contact me via PM.

3. Originally Posted by Intsecxtanx
The n−th triangular number is given by $T_{n} = \frac{1}{2}n(n+1)$. 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?
$T_{n-1}+T_n+T_{n+1} = \frac12(3n^2+3n+2)$

We need to solve the Diophantine equation $3n^2+3n+2 = 2m^2$. Let's see if you can solve this before I dive right in.

4. Diophantine equations are my weakness. I know how to tackle pell's equations with "continued fractions" but usually I don't know how else to solve them with out random guess and check methods.

5. We have $3n^2+3n+2=2m^2$

Now multiply by $12$ and add and subtract 9 to get $(6n+3)^2-24m^2+15=0$

Let $y=6n+3$ and multiply both sides by $24$ to get $(24m)^2-24y^2=360$

By letting $x=24m$ we see we have $x^2-24y^2=360$ 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 $x^2-ny^2=c$).