sum of consecutive triangular numbers is a square

Max Alekseyev maxale at
Sat Feb 24 20:17:49 CET 2007

On 2/24/07, Paulo Sachs <sachs6 at> wrote:

> If the density of such numbers is 1, it would be even
> better to have just a sequence with the numbers of
> ways in which k consecutive triangular numbers sum a
> perfect square. This sequence would be 0 in the
> 5th-10th, 12th, 14th-21st, 24th, 26th... terms. Would
> be 1 in the 32nd, 50th,... And so on. But there is
> also the possibility that there exist numbers k, such
> that there are a infinity number of ways to have k
> consecutive triangular number summing a perfect
> square.

For most k's (including k=1, k=2, k=3) the number of ways is infinite.

> By the way, Alekseyev, how did you compute those first
> terms? How can you be sure that no 5 consecutive very
> large triangular numbers add to a perfect square?

Let T(n)=n*(n+1)/2 be n-th triangular number. Then the sum of 5
consecutive terms is
T(n) + T(n+1) + T(n+2) + T(n+3) + T(n+4)
= (5*n^2 + 25*n + 40) / 2
It is a square iff 10*n^2 + 50*n + 80 is a square, i.e., the
generalized Pell equation
10*n^2 + 50*n + 80 = m^2
has non-negative solutions.

To solve such an equation, it is helpful to use Dario Alpern's applet:
The equation above has no integer solutions because it has no
solutions modulo 25.


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