# There are infinitely many triangular numbers that are the sum of two other such numbers

In the Exercise $9$, page 16, from Burton's book Elementary Number Theory he state the following:

Establish the identity $t_{x}=t_{y}+t_{z},$ ($t_{n}$ is the nth triangular numbers) where $${x}=\frac{n(n+3)}{2}+1\,\,\,\,\,\,\,y=n+1\,\,\,\,\,\,\,z=\frac{n(n+3)}{2}$$ and $n\geq 1,$ thereby proving that there are infinitely many triangular numbers that are the sum of two other such numbers.

I tried to find out how did he get $x,y$ and $z$ but I've failed. I wrote $$\frac{y(y+1)}{2}+\frac{z(z+1)}{2}=\frac{x(x+1)}{2}$$ but I don't know what to do from now on. How one can find $x,y,z$ as above?

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Multiply both sides by 8 and complete the squares, see what happens. – Will Jagy Feb 29 '12 at 23:33
@WillJagy: I've found $(2y+1)^{2}+(2z+1)^{2}=(2x+1)^{2}+1$. +1 for you. – spohreis Feb 29 '12 at 23:49
@WillJagy: I just noticed your comment. Please read this thread: meta.math.stackexchange.com/questions/1559/…. There are reasons to add an answer instead of a comment, even if it is just a hint (in this case, a substantial one and so deserves to be an answer, IMO). – Aryabhata Mar 1 '12 at 0:19
@Aryabhata, I see what you mean. In this case I felt multiplying by 8 was what I would do next, but had not the time to finish that. Also, I was a little unclear about the OP's notation. – Will Jagy Mar 1 '12 at 0:40

Triangular numbers are the sum of the first $m$ positive integers, so clearly for any $m$ there is always a triangular number which when $m$ is added to it is a triangular number: $$\frac{m(m+1)}{2}=m+\frac{m(m-1)}{2} .$$ There are others: for example any odd number greater than $1$ is the difference between two triangular numbers two steps apart, while any multiple of $3$ greater than $3$ is the difference between two triangular numbers three steps apart, etc.

So if $m$ is any triangular number, say $t_k$ where $m=\frac{k(k+1)}{2}$, then we have a triangular number which is the sum of two triangular numbers, and since there are an infinite number of triangular numbers there are an infinite number of cases of this. In this case we have $t_{m}=t_k + t_{m-1}$ or $$t_{k(k+1)/2}=t_k+t_{k(k+1)/2 - 1}.$$

Now let $k=n+1$ so $\frac{k(k+1)}{2} - 1 =\frac{(n+1)(n+2)}{2} -1 = \frac{n(n+3)}{2}$ and similarly $\frac{k(k+1)}{2}=\frac{(n+1)(n+2)}{2} = \frac{n(n+3)}{2}+1$. So the last expression of the previous result becomes $$t_{n(n+3)/2 + 1}=t_{n+1}+t_{n(n+3)/2}.$$

This explains how he got his result. It does not explain why he prefers the final step over the slightly simpler previous step.

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One way to try and come up with this would be to start from the other side:

$$\frac{y(y+1)}{2} + \frac{z(z+1)}{2} = \frac{x(x+1)}{2}$$

Multiply by $8$, and adding two gives us

$$(2y+1)^2 + (2z+1)^2 = (2x+1)^2 + 1$$

i.e.

$$(2y+1)^2 - 1 = (2x+1)^2 - (2z+1)^2$$

and so

$$y(y+1) = (x-z)(x+z+1)$$

Given a $y$, we can get a solution by putting

$$x - z = 1$$ and

$$x+z+1 = y(y+1)$$

and solving the system of equations.

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I suspect the choices come from this: start with $$(2y+1)^2 + (2z+1)^2 = (2x+1)^2 + 1.$$ If you now try to see what happens when $$x = z + 1$$ it reduces to $$(2y+1)^2 = 8 z + 9.$$ Once again, if, instead of $y$ itself, we take $$y = n+1,$$ we have $$(2 n+3)^2 = 8 z + 9$$ or $$n^2 + 3 n = 2 z$$ or $$z = \frac{n^2 + 3 n}{2}.$$

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