# Special types of Triangular numbers.

Find three consecutive triangular numbers whose sum is a perfect square.

I tried to simply use the formula that the nth triangular number is $$\frac{n(n+1)}{2}$$ and I added three conscutive triangular numbers. I tried to find cases in which the sum was a square number, but was unable to do so.

• I tried to simply use the formula that the nth triangular number is n(n+1)/2 and I added three conscutive triangular numbers. I tried to find cases in which the sum was a square number, but was unable to do so. – skallu May 14 '19 at 9:05
• The comment should be part of the question. – 5xum May 14 '19 at 9:07
• So $n(n+1)/2+(n+1)(n+2)/2+(n+2)(n+3)/2$ needs to be a square, right? – RMWGNE96 May 14 '19 at 9:13
• This simplifies to $(n+1)^2+(n+2)(n+3)/2$ and also to $n(n+1)/2+(n+2)^2$ – RMWGNE96 May 14 '19 at 9:14
• There is more than one possible answer - do you only need to find 1? – 1123581321 May 14 '19 at 9:14

HINT:

Set it up as follows

$$\frac{(n-1)n}{2} + \frac{n(n+1)}{2} + \frac{(n+1)(n+2)}{2}$$

which simplifies to

$$\frac{3n^2+3n+2}{2},$$

so you need to look at this fraction and decide for which $$n$$ is it a square.

You could also use the following two facts

• every pair of consecutive triangular numbers sums to a square

• the sum of the first $$n$$ odd numbers is equal to $$n^2$$

for another approach.

Hope this helped