Thread: Diophantine problem View Single Post
2008-05-05, 04:10   #7
wblipp

"William"
May 2003
New Haven

23×5×59 Posts

Quote:
 Originally Posted by philmoore I second Pace's suggestion, but I am still trying to understand the process of how both of you come up with these beautiful parametrized solutions. I see this sort of thing occasionally on the NMBRTHRY listserve, but I have no idea how people come up with them! Care to share any secrets?
Wouldn't they want submitters to be subscribers - or at least regular readers? I don't know how to submit it without at least reading - the web doesn't seem to offer that information to non-subscribers.

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I started with x=b, y=z=1, and hoped I could extend it. A quadratic extension would give me six "free" variables and only four constraints, so I started looking for solutions to the multiplication for

x = x1*s^2 + x2*s + b
y = y1*s^2 + y2*s + 1
z = z1*s^2 + z2*s + 1

The trickiest part was finding integer solutions for the s^4 term,
x1*y1+x1*z1+y1*z1=0

x1 = - y1 * z1 / (y1 + z1)

I did a quick search on small values of y1 and z1 looking for integer values of x1. I ran into a dead end with y1 = z1 = 2, so I tried the smallest "off diagonal" solution, y1=3 z1=6. I think Pace used this same case. I imagine other families of solutions can be made from the other off diagonal solutions.

I set x1=-2d, y1=3d, z1=6d and then looked for similar tricks to force the s^3, s^2, and s terms of the product to zero. I used the symbolic manipulation package in Mathcad (a Maple variant) at each step to express the product as a polynomial in s, then looked for a way to force one more coefficient to be zero - substitute that into the definition and repeat.

Once I had a parameterized solution to the multiplication, I made "s" a multiple of (2b+1) to easily enforce the gcd condition.

William

Last fiddled with by wblipp on 2008-05-05 at 04:21 Reason: fix a sign