Sum of five consecutive cubes is a square

[Max Alekseyev]

Oh, the sequence 2) should be excluded as all its elements are 2
modulo 3 and hence cannot be squares.

Max

On 3/10/07, Max Alekseyev <maxale at gmail.com> wrote:
> All such numbers n can be obtained from elements that are perfect
> squares of the following 4 recurrent sequence:
>
> 1) x(0)=0, x(1)=4, x(k+1) = 98*x(k) - x(k-1)
> If x(k) is a square then n = 30*x(k)
> In particular:
> for k=0, we have n=30*x(0)=0
> for k=1, we have n=30*x(1)=120
>
> 2) x(0)=-1, x(1)=11, x(k+1) = 38*x(k) - x(k-1)
> if x(k) is a square then n = 2*x(k)
>
> 3) x(0)=1, x(1)=49, x(k+1) = 38*x(k) - x(k-1)
> if x(k) is a square then n = 2*x(k)
> In particular:
> for k=1, we have n=2*x(1)=2
> for k=2, we have n=2*x(1)=98
>
> 4) x(0)=1, x(1)=9, x(k+1) = 8*x(k) - x(k-1)
> if x(k) is a square then n = 3*x(k)
> In particular:
> for k=1, we have n=3*x(1)=3
> for k=2, we have n=3*x(1)=27
>
> It also follows that for any such n one of n/2, n/3, or n/30 is a
> perfect square.
>
> I have tested 10^5 terms of each of the recurrent sequences above and
> found no new perfect squares.
>
> Max
>
> On 3/10/07, Nick Hobson <nickh at qbyte.org> wrote:
> > Hi Seqfans,
> >
> > In the Dover (2005) edition of Dickson's History of the Theory of Numbers,
> > Volume 2, Chapter 21, page 587, it is noted: "E. Lucas (275a) stated that
> > the sum of the cubes of five consecutive integers is a square only when
> > the middle number is 2, 3, 98 or 120."
> > (275a) Recherches sur l'analyse indeterminee, Moulins, 1873, 92. Extract
> >  from Bull. Soc. d'Emulation du Departement de l'Allier, 12, 1873, 532.
> >
> > The following PARI/GP script finds, additionally, middle number n = 27.
> > Is this sequence finite?  Is such a short sequence (six terms, if we
> > include n = 0) worth adding to OEIS?
> >
> > for(n=1, 10^8, if(issquare(5*n*(n*n+6)), print(n)))
> >
> > Nick
> >
>