Sum of five consecutive cubes is a square

Max Alekseyev maxale at
Sun Mar 11 01:31:22 CET 2007

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.


On 3/10/07, Nick Hobson <nickh at> 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

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