Conference rusure::math

    Let S(n) = SIGMA(i^2),i=1,n
    
    and P(n) = (1^2+2^2+...+n^2)[(n+1)^2+(n+2)^2+...+(2n)^2]
    
             = S(n) [S(2n) - S(n)]
    
                n(n+1)(2n+1)    { 2n(2n+1)(4n+1)     n(n+1)(2n+1)  }
             =  ------------  x { --------------  -  ------------  }
                     6          {       6                  6       }
    
    
    So P(n) simplifies to [n(2n+1)]^2 * (n+1)(7n+1) /36     (P an integer)
    
    P is a perfect square if (n+1)(7n+1) = m^2 is a perfect square.  
    (Sufficient but perhaps not necessary; leave that to someone else).
    
    So we need to solve equation m^2 - 7n^2 = 8n + 1 for integer solutions.
    
    Looks like a variant of Pell's equation which I don't know how to
    solve for the general solution.  Or how to convert to x^2 - Ny^2 = 1.
    
    However by inspection or otherwise, we can find n = 1, 7, 24,
    120 and 391 satisfy .0's requirements.  The values of n probably increase
    dramatically and I would guess there are infinite number of solutions.
    
    Allen        

re .1
    
>    P is a perfect square if (n+1)(7n+1) = m^2 is a perfect square.  
    
    I got this far.
    
>    (Sufficient but perhaps not necessary; leave that to someone else).
    
    It seems to me that this *is* necessary (and sufficient).
    
>    So we need to solve equation m^2 - 7n^2 = 8n + 1 for integer solutions.
    
    I don't know how to solve this but substituting t = 7n+4 leads to
    
    		t� - 7m� = 9
    
    Is that any easier? Was that a valid substitution anyway?
    
>    However by inspection or otherwise, we can find n = 1, 7, 24,
>    120 and 391 satisfy .0's requirements.
    
    Also 1921 and 6240 but I don't see any obvious pattern.
    
    
    However I tried another approach that perhaps someone can continue.
    
    Suppose (n+1)(7n+1) is a perfect square. Then there are two cases.
    
    Case I:
    
    n+1 is a perfect square and 7n+1 is a perfect square
    
    Let a� = n+1 and b� = 7n+1
    
    Therefore we have to solve 7a� - b� = 6.
    
    Case II:
    
    Neither is a perfect square ... but in this case they must be related
    in that n+1 = a�/(pqr...) and 7n+1 = (pqr...)b� where p,q,r,... are
    *distinct* primes. (There can be no repetitions otherwise we could put
    the pair inside the perfect square. Also see (*) below.)
    
    Now clearly p|a� => p|a => p�|a� => p|a�/p => p|n+1 => p|7n+7
    Clearly p|7n+1
    
    Hence p|6
    
    Hence p = 2 or p = 3
    
    So there are really three sub-cases each of which should lead to an
    equation of the form Ma� - Nb� = K. Now if only I could solve one of
    those...
    
    (*) I have ignored the possibility that n+1 = (pqr...)a� and
    7n+1 = b�/(pqr...) however by symmetry this doesn't lead to new
    solutions e.g. substitute a' = (pqr...)a etc.

 >               t� - 7m� = 9
 >      Is that any easier? Was that a valid substitution anyway?

Yes, that's valid as long as you check that this leads to integral
values of n when you substitute back.

Now that you have the equation in this form you can turn to 1837.4/.5
and get all the solutions.

In this case the generator solutions (those with m<9)
are (t,m) = (4,1) and (11,4) and the two chains of solutions are:

                             r                r
                (t) =  (8 21) (4)       (8 21) (11) 
                (m)    (3  8) (1)  and  (3  8) ( 4)


 >    Also 1921 and 6240 but I don't see any obvious pattern.

The generating process is easily converted into recurrence relations
on t and m. The one for t is:


                  t    = 16t    - t
                   r+2      r+1    r

which (since t=7n+4) yields:

                  n    = 16n    - n  + 8
                   r+2      r+1    r

(which incidentally verifies that all the n's we get are integral given
that the first two are for each chain of solutions)

This leads to the two chains of solutions for n

        0,7,120,1921,... (we can throw away 0)

        1,24,391,6240,...

Dick

>>>    P is a perfect square if (n+1)(7n+1) = m^2 is a perfect square.  
>    
>>>    (Sufficient but perhaps not necessary; leave that to someone else).
>    
>>    It seems to me that this *is* necessary (and sufficient).

I believe so but haven't proved it.  It was the /36 that made me cautious
since the factors of 2*2*3*3 might divide only part of the first part of
P(n) and part of (n+1)(7n+1).

>    
>    I don't know how to solve this but substituting t = 7n+4 leads to
>    
>                t� - 7m� = 9        (2)
>    
>    Is that any easier? Was that a valid substitution anyway?

I think so, since I got that far too.  :-)    The nasty "9" makes it hard
to solve as a standard Pell using continued fractions.

Could there be a another subsitution that converts this to x^2 - Ny^2 = 1?

Or maybe we could go back to first principles to derive the continued fraction
method for this equation. Perhaps the fact that 9 is a square may help.

                   __________
P.S.   sqrt(7)=[2, 1, 1, 1, 4]  as an infinite continued fraction.

>
>Now that you have the equation in this form you can turn to 1837.4/.5
>and get all the solutions.
>

Oops, my reply went in before I saw this!  Thanks for 1837.4/.5 solution.
I've always wanted to know how to solve x^2 - Ny^2 = M  (M not 1) but
couldn't find it in any book.

Allen

.3>This leads to the two chains of solutions for n
.3>
.3>        0,7,120,1921,... (we can throw away 0)
.3>
.3>        1,24,391,6240,...
    
    Don't throw away 0! The above two sequences are generated by the
    recurrence from .3:
    
                  a  = 16a    - a    + 8
                   n      n-1    n-2
    
    for a0=0, a1=1 (generates the bottom sequence)
    and a0=1, a1=0 (generates the top sequence)
    
            0, 1, 24, 391, 6240, 99457, ...
    
    	    1, 0,  7, 120, 1921, 30624, ...

    Kinda cute...
    
      John

.1>    So P(n) simplifies to [n(2n+1)]^2 * (n+1)(7n+1) /36     (P an integer)
.1>    
.1>    P is a perfect square if (n+1)(7n+1) = m^2 is a perfect square.  
.1>    (Sufficient but perhaps not necessary; leave that to someone else).

.4> I believe so but haven't proved it.  It was the /36 that made me cautious
.4> since the factors of 2*2*3*3 might divide only part of the first part of
.4> P(n) and part of (n+1)(7n+1).

But the first part of P(n) is a perfect square!  So whatever factors of
36 it contains, it leaves a perfect square to divide (n+1)(7n+1).
Thus, (n+1)(7n+1) must be a square.

>But the first part of P(n) is a perfect square!  So whatever factors of
>36 it contains, it leaves a perfect square to divide (n+1)(7n+1).
>Thus, (n+1)(7n+1) must be a square.

Thanks Bill.