To save people wasting excessive time on dead ends:
x^2=a+2b
Then b=1/2(x^2a)
y^2=a^2+2b^2
So y^2=a^2+2*(1/2(x^2a))^2
And 2y^2 = 3a^22ax^2+x^4
Let p=x^2: 2y^2 = 3a^22ap+p^2
This is a quadratic that can be solved for various a, with valid solutions for x, if p is itself a square = x^2.
There is a recurrence: Xn+1 = P Xn + Q Yn + K, Yn+1 = R Xn + S Yn + L. P,Q,R,S are always 3,4,2,3, K and L are both (2a), and X0 and Y0 both equal a.
So, you might think, just a matter of checking the recurrences for cases where p is square.
The only problem is that it doesn't work. It produces the first solution, a=1, b=12, x=5, y=17, and also many other solutions, none of which satisfy the requirement that a,b are relatively prime. What in fact happens is that as a gets bigger, there are variants of the solution set where XO and YO are different, though all the other values P,Q,R,S,K,L remain correct.
Say, for example, that a=257. Then 2y^2 = 3*257^22*257*p+p^2: x2  2 y2  514 x + 198147 = 0., with solutions {X0,Y)}, {257,257}, {257,257}, {15,321}, {15,321}, {211,419}, {211,419} Note that one of these solutions corresponds with Charlie's list; a=257, y=321 is indeed a solution to the given problem. But then p=15, which is not a square, even though 2*321^2 = 3*257^2+2*257*15+15^2. Working backwards, 321^2=257^2+2*b^2, so b=136. Now we have 257+2*136 = 23^2.
So somehow, the square root of 15 is 23! (In fact, 15 solves for 257+2*136, rather than 257+2*136).
So, though promisinglooking, the method doesn't produce helpful results.

Posted by broll
on 20150316 01:18:47 