In the spirit of jay_shark

[gumpzilla]

I forget if I've posted this one before, but I like it because it's a problem I posed to myself a few years ago:

Can every positive integer be represented as the ratio of two triangle numbers?

[Silent A]

LOL, I first read this as "can ANY ..."

Not sure if it's helpfull, but this is equivalent to asking:

can all positive integers be represented by: n(n+1)/[m(m+1)] where n and m are positive integers?

[jay_shark]

This is very interesting Gump .

I think you may be right .

I'll think about this one .

[Siegmund]

It's been a couple days, so I think I can post a solution without stomping on anyone's toes....

The number 4 cannot be so represented.

if it could be, we would have 4n(n+1)=m(m+1) for some integers m,n. But m=2n is too small, and m=2n+1 is too large. 4n^2+2n < 4n^2 + 4n < 4n^2 + 6n + 2 for all positive n.

Same proof works for 9.

Looks like something just slightly different will work for any perfect square.

[thylacine]

[ QUOTE ]
It's been a couple days, so I think I can post a solution without stomping on anyone's toes....

The number 4 cannot be so represented.

if it could be, we would have 4n(n+1)=m(m+1) for some integers m,n. But m=2n is too small, and m=2n+1 is too large. 4n^2+2n < 4n^2 + 4n < 4n^2 + 6n + 2 for all positive n.

Same proof works for 9.

Looks like something just slightly different will work for any perfect square.

[/ QUOTE ]

Prime factorizations lead somewhere too.

[gumpzilla]

[ QUOTE ]
It's been a couple days, so I think I can post a solution without stomping on anyone's toes....

The number 4 cannot be so represented.

if it could be, we would have 4n(n+1)=m(m+1) for some integers m,n. But m=2n is too small, and m=2n+1 is too large. 4n^2+2n < 4n^2 + 4n < 4n^2 + 6n + 2 for all positive n.

Same proof works for 9.

Looks like something just slightly different will work for any perfect square.

[/ QUOTE ]

Yep, at least on the 4. The perfect square argument sounds pretty plausible to me, too.

[Siegmund]

The perfect square argument runs onto the rocks for 36/1 and 300/3.

I misspoke - only squares of prime numbers. (The basic idea is that k^2 n (n+1) can't be rearranged to two consecutive numbers near kn since one is divisible by k and the other isn't. But for 6^2 or 10^2, it's possible to find consecutive numbers where one is a multiple of 4 and other a multiple of 9 or 25 -- 8*9/2 and 24*25/2.)

Has anyone been able to prove that it IS possible for all OTHER numbers?

[thylacine]

[ QUOTE ]
The perfect square argument runs onto the rocks for 36/1 and 300/3.

I misspoke - only squares of prime numbers. (The basic idea is that k^2 n (n+1) can't be rearranged to two consecutive numbers near kn since one is divisible by k and the other isn't. But for 6^2 or 10^2, it's possible to find consecutive numbers where one is a multiple of 4 and other a multiple of 9 or 25 -- 8*9/2 and 24*25/2.)

Has anyone been able to prove that it IS possible for all OTHER numbers?

[/ QUOTE ]

No prime powers other than 2 or 3. Consider prime factorizations of all numbers involved.