A triangular number is the sequence:

1,3,6,10,15,21,...

A square number is the sequence:

1,4,9,16,25,...

Show that there are these two sequences contain infinitely many same numbers, i.e. 1 and 1.

Results 1 to 8 of 8

- Oct 16th 2006, 07:55 AM #1

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Oct 16th 2006, 09:50 AM #2

- Oct 16th 2006, 05:26 PM #3

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Oct 17th 2006, 06:42 AM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

- Oct 17th 2006, 07:26 AM #5

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 847

- Oct 17th 2006, 08:23 AM #6

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Beautiful job Soroban +rep+

Never seen it done like that.

Your own solution?

I have modified (not really the right word) your recurrence relation into a different form:

---

I always tried to show that the cubic sequence and triangular sequence only have a trivial solution i.e. 1. And it seems to be true for higher orders also but I was unable to prove that, you know how?

- Oct 17th 2006, 01:37 PM #7

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,028
- Thanks
- 847

Hello again, TPHacker!

I got a different closed form for the recurrence.

It's not my own solution; I learned the method from a few of my books.

You may have seen something similar while exploring.

Now that LaTeX is back, I'll revise this post.

We have the recurrence: .

Assume that is exponential: .

The equation becomes: .

Divide by and we have: .

. . which has roots: .

We form a linear combination of the two roots:

. .

To solve for and , we use the first two values of the sequence:

. .

Solve the system and get: .

Therefore: .

By the way: .

. . so you can rewrite the formula if you like . . .

- Oct 23rd 2006, 08:36 AM #8

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

Besides for Soroban's solution.... I have 3 more (1 of which I am too lazy to post).

Let be the triangular number.

Solution 1)

If t_n is a square then t_{4n(n+1)} is a square. How you are suppopsed to see that, I have no idea.

Solution 2)

Following the start of Soroban;s solution we arrive at the square root of 8k² +1, which is the "Pellian equation". Thus it has infinitely many solutions.