[seqfan] Re: three triangular numbers in a row

[Max Alekseyev]

In other words, the question is when the numbers 8n+1, 16n+1, 24n+1
are squares, corresponding to the system of Diophantine equations:

8n + 1 = x^2
16n + 1 = y^2
24n + 1 = z^2

which implies

x^2 - 2*y^2 + z^2 = 0.

The parametric solution to this equation is:

x = |u^2 - 4uv + 2v^2| * p/q,
y = |u^2 - 2uv + 2v^2| * p/q,
z = |u^2 - 2v^2| * p/q,

where u,v,p,q are integers ( it can be further said that (p,q)=1 and q
equals 1, 2, or 4 - for details see my paper
http://arxiv.org/abs/1002.1679 )
with the additional constraint

2*x^2 - y^2 = 1

which gives a series of Thue equations:

(u^4 - 12*u^3*v + 32*u^2*v^2 - 24*u*v^3 + 4*v^4) * p^2 = q^2

>From (p,q)=1, we have p=1 and thus there are just four Thue equations
corresponding q=1, 2, and 4.
It is easy to find all their solutions in PARI/GP:

? t=thueinit(x^4 - 12*x^3 + 32*x^2 - 24*x + 4);
? thue(t,1^2)
%1 = [[1, 1], [-1, -1], [1, 0], [-1, 0]]
? thue(t,2^2)
%2 = [[0, 1], [0, -1], [2, 1], [-2, -1]]
? thue(t,4^2)
%3 = [[2, 2], [-2, -2], [2, 0], [-2, 0]]

All these solutions actually give just n = 0.

So n, 2n, 3n are triangular only for n=0.

Regards,
Max


On Fri, Jun 24, 2011 at 7:58 AM, Brendan McKay <bdm at cs.anu.edu.au> wrote:
> A question related to A075528 and A061278 was asked at
>   http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Striped_triangle_question
>
> It is equivalent to:
>   Is there any n such that n, 2n, and 3n are all triangular numbers?
>
> Brendan.
>
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