A Diophantine equation question?

[Maximilian Hasler]

Let c(m) = (m-1)³+m³+(m+1)³ = 3m(m²+2)
If c(m) is a square, then 3 | m(m²+2).
since 3 | (m²+2) is impossible (-1 is not a square mod 3),
m=3n.
c(m) = c(3n) = 9n(9n²+2) is a square if  n(9n²+2) is a square.
let n=ab² with a squarefree.
c(m) is a square if a(9a²b^4+2) is a square
=> a | (9a²b^4+2)
=> a | 2
=> a = 1 or 2
a=1 => c(m) = 9b²(9b^4+2) is a square if  (9b^4+2) is a square, which
is impossible.
a=2 => c(m) = 9*2*b²(9*4*b^4+2) = 9*4*b²(9*2*b^4+1)
The solutions to 18x²+1=y² are
[xn,yn] = [17, 4; 72, 17 ]^n * [0,1]

I get : x[n] = 4 * A029547[n-1]

Thus, the question is : is there an element in the sequence
A029547 other than a(0)=1 which is a square ?
Any squares in this sequence correspond to solutions m[n] = 6 x[n].

M.H.


On 5/26/07, Jon Schoenfield <jonscho at hiwaay.net> wrote:
> A couple of weeks ago, Neil forwarded to the list a set of several questions
> I had asked regarding A116108 and a related sequence.  My main question was
> (and still is) whether A116108 is finite and, if so, whether 41616 is the
> final term.  Could anyone shed some light on this?
>
> Thanks,
>
> -- Jon
>
>
>