I apologize for derailing the OP's birthday thread!

I'll go read up some more on triangular numbers. My first impulse is to write an ugly Perl program to brute force something but instead I'll try to understand it mathematically.

I knew I shouldn't have trusted Wikipedia. (I picked up my original claim there. Every year I check my age to see what interesting things are associated with that number. I should have checked before posting!)

:newcat:

[QUOTE=Mini-Geek;89196]Because it's his age? Unless he's over 170[/QUOTE]

I checked up to 10[sup]250[/sup], so I already knew 36 was the only reasonable age. But this wasn't stated as a puzzle, it was stated as a mathematical fact.

[QUOTE=Mini-Geek;89196]due to numbers being infinite, I would think that there are likely many more[/QUOTE]

I have a heuristic that suggests there are no more. But I was wondering if somebody had a proof.

It is fairly easy to show this:

Let n(n+1)/2 = [m(m+1)/2]^2 = m^2(m+1)^2/4

i.e., n(n+1) = m^2(m+1)^2/2

But since n and (n+1) are relatively prime, we must have either

n=m^2, n+1 = (m+1)^2/2
or
n=(m+1)^2/2, n+1 = m^2
[I'll leave it as an excercise why the /2 factor must go with (m+1)^2:wink:]

The first one leads to m=1 (solution = 1, 1), the second one leads to m=3 (solution = 6, 36)

LOL, come to think of it, we're so geeky at times. :wink:

[QUOTE=axn1;89200]It is fairly easy to show this:

Let n(n+1)/2 = [m(m+1)/2]^2 = m^2(m+1)^2/4

i.e., n(n+1) = m^2(m+1)^2/2

But since n and (n+1) are relatively prime, we must have either

n=m^2, n+1 = (m+1)^2/2
or
n=(m+1)^2/2, n+1 = m^2[/QUOTE]

I don't see why you can reject the possibility that

m = pq
m+1 = st

2n = p[sup]2[/sup]s[sup]2[/sup]
n+1 = q[sup]2[/sup]t[sup]2[/sup]

OR

n = p[sup]2[/sup]s[sup]2[/sup]
2(n+1) = q[sup]2[/sup]t[sup]2[/sup]

[QUOTE=wblipp;89209]I don't see why you can reject the possibility that

m = pq
m+1 = st

2n = p[sup]2[/sup]s[sup]2[/sup]
n+1 = q[sup]2[/sup]t[sup]2[/sup]

OR

n = p[sup]2[/sup]s[sup]2[/sup]
2(n+1) = q[sup]2[/sup]t[sup]2[/sup][/QUOTE]

Hmmm... Didn't think of that. Any chance of salvaging this approach, however? Some restrictions on sizes, etc?

[quote=Xyzzy]I knew I shouldn't have trusted Wikipedia.[/quote]

Mathworld.

[URL]http://mathworld.wolfram.com/[/URL]

The self-proclaimed "Web's Most Extensive Mathematical Resource".

Remember it in your time of need.

(now that Eric Weisstein and CRC Press have settled their legal dispute about CRC Press's shameful (IMO) power play on Eric's naivete a few years ago -- see [URL]http://mathworld.wolfram.com/docs/faq.html[/URL])

- - -

From [URL]http://mathworld.wolfram.com/TriangularNumber.html[/URL]:

"The numbers 1, 36, 1225, 41616, 1413721, 48024900, ... (Sloane's [URL="http://www.research.att.com/~njas/sequences/A001110"][COLOR=#800080]A001110[/COLOR][/URL]) are [URL="http://mathworld.wolfram.com/SquareTriangularNumber.html"][COLOR=#0000ff]square triangular numbers[/COLOR][/URL], i.e., numbers which are simultaneously triangular and [URL="http://mathworld.wolfram.com/SquareNumber.html"][COLOR=#0000ff]square[/COLOR][/URL] (Pietenpol 1962). The corresponding square roots are 1, 6, 35, 204, 1189, 6930, ... (Sloane's [URL="http://www.research.att.com/~njas/sequences/A001109"][COLOR=#800080]A001109[/COLOR][/URL]), and the indices of the corresponding triangular numbers [T[sub]n[/sub]] are [n=1], 8, 49, 288, 1681, ... (Sloane's [URL="http://www.research.att.com/~njas/sequences/A001108"][COLOR=#0000ff]A001108[/COLOR][/URL])."

So, 1 and 36 are not the only possibilities, but Xyzzy'd have to be older than Methuselah otherwise.

[QUOTE=Xyzzy;89162]How embarassing! I wonder how I should reword this? Let me go think for a while.[/QUOTE]Ok, I've waited long enough for you to get your act together.

The smallest triangular number which is the square of a different triangular number.


Paul

[QUOTE=cheesehead;89403]The corresponding square roots are 1, 6, 35, 204, 1189, 6930,
....
So, 1 and 36 are not the only possibilities, but Xyzzy'd have to be older than Methuselah otherwise.[/QUOTE]And which of those square roots are themselves triangular?



Paul

[quote=xilman;89410]The smallest triangular number which is the square of a different triangular number.[/quote]
That'll do :piggie:. That'll do.

At this point I'm just waiting for my birthday so I can put all this behind me.

:whistle:

[QUOTE=xilman;89411]And which of those square roots are themselves triangular?[/QUOTE]
Yes, cheesehead seems to have forgotten that the square root must also be a triangular number. As I reported in post #13, I've checked that there are no other triangular square roots up to 10[sup]250[/sup] (the square root this large, so the number up to 10[sup]500[/sup]). I don't have a proof there are no others, though.