Thread: Diophantine Question View Single Post
2009-09-16, 21:04   #7
maxal

Feb 2005

22·32·7 Posts

Quote:
 Originally Posted by grandpascorpion Background: I'm trying to create an a.p. of 6 or more terms. (http://www.primepuzzles.net/puzzles/puzz_413.htm)
That's a tough problem.
If $n$ is such that for some $k$ of its divisors: $d_1, d_2, \dots, d_k$, we have
$\frac{n}{d_i} + d_i = m + q\cdot i$ for $i=1,2,\dots,k$
then
$(m+qi)^2 - 4n = \left( \frac{n}{d_i} - d_i \right)^2$ for $i=1,2,\dots,k$
form a sequence of $k$ squares whose second differences equal the constant $2 q^2$.

For example, $n=36400$ gives a sequence of squares
$33^2, 150^2, 213^2, 264^2, 309^2$
whose second differences equal $2\cdot 27^2 = 1458$.

Finding sequences of squares with constant second differences is a rather hard task (see the attached paper) and additional requirement of having difference of the special form $2 q^2$ makes it even harder.
Attached Files
 browkin8572.pdf (146.3 KB, 245 views)

Last fiddled with by maxal on 2009-09-16 at 21:24