From the equation, we know that sē must be odd, since the term
(4p4+ 4p3+ 4p2 + 4p) is even. Therefore, s must also be odd.
We also know that for every odd s:
(s+1)(s-1) is evenly divisible by 4.
The obvious solutions are:
{(0,-1), (0,1), (2,-11), (2,11)}However, as
p increases, something ppeculiar happens.
√
(4p4+ 4p3+ 4p2 + 4p) = i + r
where i = an integer &
r = the remainder. The solution of course is to find a
p where
r = 0. However, as
p→∞, it appears that
r→0.75. If this holds true, then there are no more solutions for this equation.
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Posted by hoodat
on 2007-05-14 13:25:19 |
Partial solution
