The Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Pell type equations
Replies: 29   Last Post: Apr 29, 2012 5:37 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Timothy Murphy

Posts: 657
Registered: 12/18/07
Re: Pell type equations
Posted: Apr 28, 2012 8:39 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Terry M wrote:

> The reason I was hoping there was a way to plug this type of equation into
> Brahmagupta's method of composition is that I can find many solutions for
>
> 5x^2 + 20 = y^2
>
> where x and y are coprime but can find no such solutions for 5x^2 + 45 =
> y^2


As before, 5 | y, and writing y = 5z we get

x^2 - 5z^2 = -9.

An equation of the form x^2 - dy^2 = c
(where d is not a perfect square)
may have no solution;
but if you can find one you can find an infinite number
by combining this solution with the general solution
of Pell's equation x^2 - dy^2 = 1
(which always has an infinity of solutions),
in the way I suggested.

The equation x^2 - dy^2 = c has a solution
if it has a solution modulo 8d, I think.

In your case this means there must be solutions mod 8 and mod 5,
which there are.

Actually, in this case it is sufficient to find a solution of

u^2 - 5v^2 = -1

since then x = 3u, z = 3v
(and it's not difficult to see that every solution
must be of this form, ie x and z must be divisible by 3).

This has the trivial solution (u,v) = (2,1),
or (x,z) = (6,3) or (x,y) = (6,15)

So there are an infinity of solutions,
which you can get in the way I suggested.
Eg if e = 2 + sqrt5 then e^3 = 38 + 17 sqrt5,
so (u,v) = (38,17) is a solution of u^2 - 5y^2 = -1,
giving (3 x 38, 15 x 17) as a solution of the original equation.


--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College Dublin




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.