21
Shubhodip
·2011-01-28 01:42:17
my solution : if the given expression equals zero (p,p2) , (q,q2) , (r,r2) will be collinear
p,p2 ,q,q2 ,r,r2 are distinct points on the parabola y= x2
but a straight line intersects a parabola at maximum 2 points
so the given is nonzero
i am looking for a basic algebric proof
71
Vivek @ Born this Way
·2011-01-28 01:52:07
(p,p2) , (q,q2) , (r,r2)
How do you infere this please let me know.
341
Hari Shankar
·2011-01-28 02:51:52
Its true for the simple reason that \sum pq(p-q) = -(p-q)(q-r)(r-p)
Now, how did we arrive at the factorisation?
This is a well known technique. By setting p=q, we see that the expression equals zero
So, (p-q) must be a factor of LHS.
So (p-q)(q-r)(r-p) divides LHS. Comparing the degree, we see that
LHS = K (p-q)(q-r)(r-p) where K is a constant.
Comparing the coeff of p on both sides, K=-1 and we are done
21
Shubhodip
·2011-01-28 05:31:48
thank you very much sir thanks a lot
341
Hari Shankar
·2011-01-28 06:56:47
We can tie this with Shubhodip's remark about the points.
If the points are collinear then the are of the triangle formed by the three points is zero i.e.
\begin{vmatrix} 1 & p& p^2 \\ 1&q &q^2 \\ 1 & r & r^2 \end{vmatrix} = 0
Now look at the Vandermonde determinant
http://en.wikipedia.org/wiki/Vandermonde_matrix
You can now see why we must have p≠q≠r