0
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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. $\endgroup$ – skallu May 14 '19 at 9:05
  • $\begingroup$ The comment should be part of the question. $\endgroup$ – 5xum May 14 '19 at 9:07
  • $\begingroup$ So $n(n+1)/2+(n+1)(n+2)/2+(n+2)(n+3)/2$ needs to be a square, right? $\endgroup$ – RMWGNE96 May 14 '19 at 9:13
  • $\begingroup$ This simplifies to $(n+1)^2+(n+2)(n+3)/2$ and also to $n(n+1)/2+(n+2)^2$ $\endgroup$ – RMWGNE96 May 14 '19 at 9:14
  • $\begingroup$ There is more than one possible answer - do you only need to find 1? $\endgroup$ – 1123581321 May 14 '19 at 9:14
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.