A Diophantine equation question?

[Maximilian Hasler]

Sorry for the delay of my reply, I had some internet connection problems.
I completed my simple analysis for the case where m is not a multiple of 3.
If I did not make another error,  I can assert :

* The next term, if it exists, is (much) larger than 10^150000.

* Elements of this sequence are of the form
	c(m) = (m-1)^3+m^3+(m+1)^3 = 3m(m^2+2) with
 - either m = 24 x, where  x  are the squares in A091761 = (0, 1, 34, 1155,...)
 - or m is not a multiple of 3 and
 either any (odd) square in the sequence A001834 = (1, 5, 19, 71, ...)
 or m = 2x with x any square in the sequence A054320= (1, 11, 109, 1079,...).

For any of the 3 cases, no element >1 of the mentioned sequences seems
to be a square. Thus, it appears that the 4 known terms are the only
possible ones.
My reasoning follows.

c(m) = 3m(m²+2) = (m-1)^3 + m^3 + (m+1)^3
is a square if 3 | m(m²+2) <=> (a) m=3n or (b) (m²+2)=3n
(a) m=3n : see my prevoius mail : we must have m=24*( a square of the
sequence A091761 ).

(b) Let  m=ab²  with  gcd(3,a)=gcd(3,b)=1,  a  square free.
    c(m) = 3ab²(a²b^4+2)  is a square  <=>  3a | a²b^4+2  (since not 3|a).
    	<=>   a | a²b^4+2  (since 3 | a²b^4+2  always true for gcd(ab,3)=1)
    	<=>   a | 2   <=>   a=1 or a=2.
    (i) a=1,  c(m) = 3b²(b^4+2) is a square iff  b^4+2 = 3n²
    	Let x=b², => x²+2 = 3n²
    	  	 <=>  x = x[k] = 4 x[k-1] - x[k-2], x[0]=1=-x[-1]
    			= ( 1, 5, 19, 71, ...) = A001834

    	Maybe for some seqfan it is not difficult to show that there is
    	no square in this sequence except for x[0]=1.
    	At least, I checked that there is none below x[10^5] ~ 10^57200
(PARI:)
fx1(n=10^5,x=[1,-1],A=[4,1;-1,0])=while(!issquare((x*=A)[1]) &
n--,);if(n,x[1],log(x[1])/log(10))

   (ii) a=2,  c(m) = 6b²(4b^4+2) = 12b²(2b^4+1) is a square iff
   	2b^4+1 = 3n². 2x²+1 = 3n² iff  x in (1, 11, 109, 1079,...) = A054320
   	i.e. in the sequence  x[k] = 10x[k-1]-x[k-2], x[0]=1=-x[-1].

I checked that there is no square >1 in this sequence up to x[10^5] ~ 10^99559.

(PARI:)
fx2(n=10^5,x=[1,-1],A=[10,1;-1,0])=while(!issquare((x*=A)[1]) &
n--,);if(n,x[1],log(x[1])/log(10))

Thanks in advance for any new bug reports...
M.H.

On 5/26/07, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> Oh sorry, of course (since -2 = 1 IS a square mod 3...
> this gives b.t.w. the term 9 in the sequence, for m=1).
> So the interesting connection with A029547 applies only to the
> subsequence of squares c(m=3n).
> Maximilian
>
> On 5/26/07, Don Reble <djr at nk.ca> wrote:
> > > since 3 | (m²+2) is impossible
> >    Eh?  3 | (1^2+2), 3 | (2^2+2), 3 | (4^2+2), ...
> >    3 divides, unless m is a multiple of 3.
> > --
> > Don Reble  djr at nk.ca
> >
> > --
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>