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

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

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

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

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

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

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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

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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: . $\displaystyle f(n) \:=\:6\cdot f(n-1) - f(n-2)$

Assume that $\displaystyle f(n)$ is exponential: . $\displaystyle f(n) \:=\:X^n$

The equation becomes: . $\displaystyle X^n \:=\:6X^{n-1} - X^{n-2}$

Divide by $\displaystyle X^{n-2}$ and we have: . $\displaystyle X^2 - 6X + 1 \:= \:0$

. . which has roots: .$\displaystyle X \:=\:3 \pm 2\sqrt{2}$

We form a linear combination of the two roots:

. . $\displaystyle f(n) \:=\:A(3 + 2\sqrt{2})^n + B(3 - 2\sqrt{2})^n$

To solve for $\displaystyle A$ and $\displaystyle B$, we use the first two values of the sequence:

. . $\displaystyle \begin{array}{cc}f(1) \:= \\ f(2) \:=\end{array} \begin{array}{cc} A(3 + 2\sqrt{2}) + (3 - 2\sqrt{2})\\ A(3 + 2\sqrt{2})^2 + (3 - 2\sqrt{2})^2\end{array}\begin{array}{cc}= \:1 \\ = \:6\end{array}$

Solve the system and get: . $\displaystyle A \,= \,\frac{1}{4\sqrt{2}},\;\;B \,= \,-\frac{1}{4\sqrt{2}}$

Therefore: . $\displaystyle f(n) \;= \;\frac{(3 + 2\sqrt{2})^n - (3 - 2\sqrt{2})^n}{4\sqrt{2}} $

By the way: . $\displaystyle 3 \pm 2\sqrt{2} \:=\:(1 \pm \sqrt{2})^2$

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

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

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

Let $\displaystyle t_n$ be the $\displaystyle n-th$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.