A problem

**namelessguy** · December 7th, 2008, 09:14 PM

I'm having difficulty with proving this problem. Can someone give some help?
Prove that the sum of the first $n$ natural numbers is a perfect square for infinitely many values of $n$ .

I don't really know if I interpret this question correctly. I know that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$ . But this is not in a form a perfect square, isn't it? I'm really confused on this.

**NonCommAlg** · December 7th, 2008, 09:59 PM

Originally Posted by namelessguy

I'm having difficulty with proving this problem. Can someone give some help?
Prove that the sum of the first $n$ natural numbers is a perfect square for infinitely many values of $n$ .

I don't really know if I interpret this question correctly. I know that the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$ . But this is not in a form a perfect square, isn't it? I'm really confused on this.

it's quite simple if you're familiar with Pell equation (i think everybody is!) anyway, so we want to have $n(n+1)=2k^2$ for some integer $k.$ equivalently $(2n+1)^2 - 2(2k)^2 = 1,$ which is a Pell

equation. thus if we let $A= \left \{\frac{(3+2\sqrt{2})^m + (3-2\sqrt{2})^m - 2}{4}: \ \ m \in \mathbb{N} \right \},$ then for any $n \in A$ the sum $1+2+ \cdots + n$ will be a perfect square. note that $A \subset \mathbb{N}. \ \Box$

**namelessguy** · December 7th, 2008, 10:28 PM

Originally Posted by NonCommAlg

it's quite simple if you're familiar with Pell equation (i think everybody is!) anyway, so we want to have $n(n+1)=2k^2$ for some integer $k.$ equivalently $(2n+1)^2 - 2(2k)^2 = 1,$ which is a Pell

equation. thus if we let $A= \left \{\frac{(3+2\sqrt{2})^m + (3-2\sqrt{2})^m - 2}{4}: \ \ m \in \mathbb{N} \right \},$ then for any $n \in A$ the sum $1+2+ \cdots + n$ will be a perfect square. note that $A \subset \mathbb{N}. \ \Box$

Thank you very much for your help, NonCommAlg. I just learned Pell equation. I don't know if I follow you correctly on this $n(n+1)=2k^2 \Rightarrow 2n^2+2n=2(2k^2)$ $\Rightarrow 2n^2+2n-2(2k^2)=0 \Rightarrow 2n^2+2n+1-2(2k^2)=1$
I can't get $(2n+1)^2-2(2k^2)=1$ as yours. Can you help me figure out what I did wrong here?

Edit: I see what I did wrong, instead of $(2k)^2$ , I put $2k^2$

**ThePerfectHacker** · December 8th, 2008, 05:18 AM

If $t_n$ is a square then $t_{4n(n+1)}$ is a square.

**namelessguy** · December 9th, 2008, 10:58 PM

Can you show me how you come up with that set $A$ NonComAlg. I know the terms $3+2\sqrt{2}$ is from the least positive solution x=3,y=2

**ThePerfectHacker** · December 10th, 2008, 11:19 AM

Originally Posted by namelessguy

Can you show me how you come up with that set $A$ NonComAlg. I know the terms $3+2\sqrt{2}$ is from the least positive solution x=3,y=2

One way is knowning how to solve Pell's equation.
I can show you the steps but if you never learned how to solve Pell's equation then it will be unhelpful.

**namelessguy** · December 10th, 2008, 02:59 PM

Originally Posted by ThePerfectHacker

One way is knowning how to solve Pell's equation.
I can show you the steps but if you never learned how to solve Pell's equation then it will be unhelpful.

I know how to find the least positive solution to Pell's equation using the continued faction expansion of $\sqrt{d}$ . Then all the solutions are $(x_n,y_n)$ by $x_n +\sqrt{d} y_n= (x_1+ \sqrt{d} y_1)^n$ where $(x_1,y_1)$ is the least positive solution.
I don't know what I need to do with the solution to come up with that set $A$ , so if you can give some help, it would be great, TPH. Thanks a lot.

**Soroban** · December 12th, 2008, 06:03 PM

Hello, namelessguy!

NonCommAlg solved the Pellian equation
. . and gave us the closed form of the solutions.

The solution is also a recurrence relation . . .

We want to find integers $a\text{ and }b$ so that: . $\frac{a(a+1)}{2} \:=\:b^2$

By experimenting, we can find the first few values . . .

. . $\begin{array}{c|c} a & b \\ \hline<br /> 0 & 0 \\ 1 & 1 \\ 8 & 6 \\ 49 & 35 \\ \vdots & \vdots \end{array}$

Then we can generalize the two sequences:

. . $a_n \;=\;6\!\cdot\!a_{n\,\text{-}\,1} - a_{n\,\text{-}\,2} + 2$

. . $b_n \;=\;6\!\cdot\!b_{n\,\text{-}\,1} - b_{n\,\text{-}\,2}$

And extend the table indefinitely:

. . $\begin{array}{c|c} a & b \\ \hline<br /> \vdots & \vdots \\ 288 & 204 \\ 1681 & 1189 \\ 9800 & 6930 \\ \vdots & \vdots \end{array}$

Therefore, there is an infinite number of square triangular numbers.

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