Find all (n,m)
Kenshin
n^2 = m^2.
I find (1,1) and (24,70).
However, is there anything else?
I've tried to use the equality 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
and the fact that n, n+1, 2n+1 are relatively prime but there's
nothing to be really helpful.
Any kind of hint?
Thank you for your advice.
Best wishes.
Kenshin.
christian.bau
> Find all (n,m) (pair of natural number) such that 1^2 + 2^2 + ... +
> n^2 = m^2.
>
> I find (1,1) and (24,70).
>
> However, is there anything else?
>
> I've tried to use the equality 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
> and the fact that n, n+1, 2n+1 are relatively prime but there's
> nothing to be really helpful.
>
> Any kind of hint?
Since they are relative prime, they must have the form 6x^2, y^2, z^2,
or 2x^2, 3y^2, z^2.
For example, n = 24: 24 = 6 * 2^2, n+1 = 25 = 5^2, 2n+1 = 49 = 7^2.
There's the question which of n, n+1, 2n+1 has which coefficient. So
you can set up a few different equations. Each will be a diophantine
equation. Each will have a very small number of solutions. For
example, the solutions of (n+1 is square, 2n+1 is square) are related
to the continued fraction for sqrt (2).
For example: To find all x where 6 x^2 + 1 = y^2 (like x = 2, y = 5,
giving n = 24 and n+1 = 25), write a computer program to find the
first few solutions say up to a million, then find a pattern, and you
will be able to find all solutions quite quickly. Then you check for
which of these 2n+1 is also a square. Same for other combinations,
like 3x^2 + 1 = y^2. If you Google for it you will find a solver for
these on the internet.
The Pumpster
This is Lucas' square pyramid problem. There are no other solutions
in positive integers. The original proof is due to Watson; more
recently, there have been found a number of elementary proofs, by Ma,
Anglin and others.
de P
Kenshin
Gerry Myerson
<98c28c66-f361-4a6a...@v31g2000vbs.googlegroups.com>,
"christian.bau" <christ...@cbau.wanadoo.co.uk> wrote:
> On Feb 11, 5:17 pm, Kenshin <rurouni sohj...@hanmail.net> wrote:
> > Find all (n,m) (pair of natural number) such that 1^2 + 2^2 + ... +
> > n^2 = m^2.
> >
> > I find (1,1) and (24,70).
> >
> > However, is there anything else?
> >
> > I've tried to use the equality 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
> > and the fact that n, n+1, 2n+1 are relatively prime but there's
> > nothing to be really helpful.
> >
> > Any kind of hint?
>
> Since they are relative prime, they must have the form 6x^2, y^2, z^2,
> or 2x^2, 3y^2, z^2.
>
> For example, n = 24: 24 = 6 * 2^2, n+1 = 25 = 5^2, 2n+1 = 49 = 7^2.
>
> There's the question which of n, n+1, 2n+1 has which coefficient. So
> you can set up a few different equations. Each will be a diophantine
> equation. Each will have a very small number of solutions. For
> example, the solutions of (n+1 is square, 2n+1 is square) are related
> to the continued fraction for sqrt (2).
You're not saying that n + 1 = x^2, 2 n + 1 = y^2
has a very small number of solutions, I'm sure
(but it reads as if you were saying that).
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)