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.

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- October 16th 2006, 07:55 AMThePerfectHackerQuestion 2
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. - October 16th 2006, 09:50 AMtopsquark
- October 16th 2006, 05:26 PMThePerfectHacker
- October 17th 2006, 06:42 AMThePerfectHacker
I do not understand the last line of the solution?

- October 17th 2006, 07:26 AMSoroban
Hello, TPHacker!

Generally, with this type of problem we can generate solutions

with a recurrence formula of the form:

. .*n*th term .= .k·(preceding term) ± (term before) ... for some constant*k.*

In short, I "eyeballed" it . . .

- October 17th 2006, 08:23 AMThePerfectHacker
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? - October 17th 2006, 01:37 PMSoroban
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 . . .

- October 23rd 2006, 08:36 AMThePerfectHacker
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.