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 triangularsquare numbers? Do you think that there are infinitely many? This question is from http://www.amazon.com/FriendlyIntro.../dp/0321816196. I'm totally new to number theory. Of course, I googleed 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 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
March 10th, 2015, 06:30 AM  #5  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Quote:
$\displaystyle u_{n+2}6u_{n+1}+u_{n}=0$ You find the characteristic roots are: $\displaystyle r=3\pm2\sqrt{2}$ And so the general form of the closed solution is: $\displaystyle u_{n}=k_1\left(3+2\sqrt{2}\right)^n+k_2\left(32\sqrt{2}\right)^n$ Using the initial conditions, we find: $\displaystyle u_0=k_1+k_2=0$ $\displaystyle u_1=k_1\left(3+2\sqrt{2}\right)+k_2\left(32\sqrt{2}\right)=1$ From this, we find: $\displaystyle \left(k_1,k_2\right)=\left(\frac{1}{4\sqrt{2}},\frac{1}{4\sqrt{2}}\right)$ Hence: $\displaystyle u_{n}= \frac{1}{4\sqrt{2}}\left(\left(3+2\sqrt{2}\right)^ n \left(32\sqrt{2}\right)^n\right)$  
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$$T_n = \tfrac12 n (n1) = \begin{cases} k(2k + 1) &\text{or} \\ k(2k  1)\end{cases}$$ 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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number, square, triangular 
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