I'm 21 now!
 

Well, that was three days ago. :P

The cool thing is that my age is now the product of the first two Mersenne primes! :D

[QUOTE=ixfd64;89105]Well, that was three days ago. :P

The cool thing is that my age is now the product of the first two Mersenne primes! :D[/QUOTE]

... which are the digits of my age! :cool:

[QUOTE=alpertron;89108]... which are the digits of my age! :cool:[/QUOTE]I didn't realise you were 73. You're probably the oldest person here! :wink:

My age is the square of a Mersenne prime, and I'm not unusually precocious.


Paul

When you get to my age you will understand the answer to life, the universe and everything.

[QUOTE=ixfd64;89105]I'm 21 now![/QUOTE]Congratulations !
I'm 48 since yesterday Friday 13th. Half of my life :wink: . But I bet I've already lived the best half yet.
Enjoy this preriod of your life !
BTW, I'm not sure California is the best place to live long and happy: stress, pollution, too-fat and too-sweety meals, ... Your opinion ?
Tony

[quote=Flatlander;89127]When you get to my age you will understand the answer to life, the universe and everything.[/quote]That's easy: 42.

My age is the only triangular number whose square root is also a triangular number.

[QUOTE=Xyzzy;89138]My age is the only triangular number whose square root is also a triangular number.[/QUOTE]I flatly refuse to believe you are only one year old.

Paul

How embarassing! I wonder how I should reword this? Let me go think for a while.

:poop:


[SIZE=1][I]Edit #1: My age is the largest triangular number whose square root is also a triangular number? Crap, I'm in over my head now!

Edit #2: Well, if mental ability, bowel control and hairiness are indicators of age, I may well be 1!
[/I][/SIZE]

[QUOTE=Xyzzy;89138]My age is the only triangular number whose square root is also a triangular number.[/QUOTE]
How do you prove that 1 and 36 (and zero if you choose to admit it as a triangle number) are the only numbers with this property?

I was able to show there are an infinite number of triangle numbers that are perfect squares - I was even able to generate them via a recurrance relationship. But how do you prove that none except these early ones are squares of triangle numbers?

[quote=wblipp;89179]How do you prove that 1 and 36 (and zero if you choose to admit it as a triangle number) are the only numbers with this property?

I was able to show there are an infinite number of triangle numbers that are perfect squares - I was even able to generate them via a recurrance relationship. But how do you prove that none except these early ones are squares of triangle numbers?[/quote]Because it's his age? Unless he's over 170 (that's up to where I checked the triangle number's sq roots), or is 1 year old and is somehow able to type and understand what triangle numbers and square roots are, he's 36.

And no, I don't know if 1 and 36 are the only triangle numbers with that property (in fact, due to numbers being infinite, I would think that there are likely many more), but they're the only ones with that property that are possible (with current medical technology, of course) to be an age.

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.

[quote=xilman;89411]And which of those square roots are themselves triangular?[/quote]
[quote=wblipp]Yes, cheesehead seems to have forgotten that the square root must also be a triangular number.[/quote]I didn't forget, but I did misunderstand MathWorld's reference to A001110, and I misread 36 in A001109 as "35", which gave pretty much the same effect. :blush:

MathWorld's your friend, but sometimes you need better glasses to appreciate a friend. I'm putting that on my to-do list.