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 March 9th, 2015, 10:03 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Square triangular number The first two numbers that are both squares and triangles are 1 and 36. Find the next one and, if possible, the one after that. Can you figure out an efficient way to find triangular-square numbers? Do you think that there are infinitely many? This question is from http://www.amazon.com/Friendly-Intro.../dp/0321816196. I'm totally new to number theory. Of course, I google-ed this question and I got the answer as well. But it's useless for me because I don't know how to generate the formula to find the answer? Can anyone explain how to figure out an efficient way to find the numbers ? Thanks.
 March 10th, 2015, 04:58 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Have you seen this page: Square Triangular Number -- from Wolfram MathWorld?
 March 10th, 2015, 05:03 AM #3 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Yes. I just learned the general formula. How to derive the general formula actually ?
March 10th, 2015, 05:23 AM   #4
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Quote:
 Originally Posted by jiasyuen Yes. I just learned the general formula. How to derive the general formula actually ?
You can use Pell's equation, I believe.

March 10th, 2015, 06:30 AM   #5
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Quote:
 Originally Posted by jiasyuen Yes. I just learned the general formula. How to derive the general formula actually ?
If you take the 2nd order recurrence:

$\displaystyle u_{n+2}-6u_{n+1}+u_{n}=0$

You find the characteristic roots are:

$\displaystyle r=3\pm2\sqrt{2}$

And so the general form of the closed solution is:

$\displaystyle u_{n}=k_1\left(3+2\sqrt{2}\right)^n+k_2\left(3-2\sqrt{2}\right)^n$

Using the initial conditions, we find:

$\displaystyle u_0=k_1+k_2=0$

$\displaystyle u_1=k_1\left(3+2\sqrt{2}\right)+k_2\left(3-2\sqrt{2}\right)=1$

From this, we find:

$\displaystyle \left(k_1,k_2\right)=\left(\frac{1}{4\sqrt{2}},-\frac{1}{4\sqrt{2}}\right)$

Hence:

$\displaystyle u_{n}= \frac{1}{4\sqrt{2}}\left(\left(3+2\sqrt{2}\right)^ n- \left(3-2\sqrt{2}\right)^n\right)$

 March 10th, 2015, 02:57 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra The $n$th triangle number $T_n$ is given by$$T_n = \tfrac12 n (n-1) = \begin{cases} k(2k + 1) &\text{or} \\ k(2k - 1)\end{cases}$$ for some natural number $k$. I shall henceforth write $T_n = k(2k \pm 1)$ to represent this. Now, if a natural number $r \gt 1$ divides $k$, then $r$ also divides $2k$. Thus $r$ does not divide $(2k \pm 1)$. We want $T_n$ to be square, so $r \divides T_n \implies r^2 \divides T_n$ and thus $r^2$ also divides $k$. Hence $k$ is square. Since both $T_n$ and $k$ we must have that $2k \pm 1)$ is also square. Thus, to find square triangular numbers, examine $(2k \pm 1)$ for each square $k$. If either $(2k \pm 1)$ is also square then the triangular number $k(2k \pm 1)$ is a square triangle number.

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