Not as straightforward as it looks!
Let (y^2-1)/(x^2-y^2+1) = k, some integer multiple, and let (x^2-y^2 + 1) = b.
1. Consider the fundamental solution for each successive k, with x and y differing by 1:
(y^2-1)/(x^2-(y^2-1)) = 1
{b == 8, x == 4, y == 3},
(y^2-1)/(x^2-(y^2-1)) = 2
{b == 12, x == 6, y == 5}
etc. with x passing over the even integers, and y over the odd ones.
(y^2-1)/((y+1)^2-(y^2-1)) =(y-1)/2; so y is odd, and x is even. This is true for all k, and b can be square, if k is, e.g. (y^2-1)/(x^2-(y^2-1)) = 3; {b == 16, x == 8, y == 7}.
2. There is one exception. If k is one less than a square, additional fundamental solutions are then possible, with x=y;
(y^2-1)/(y^2-(y^2-1))=(y^2-1) = k:
(y^2-1)/(x^2-(y^2-1)) = 3
{b == 1, x == 2, y == 2}
(y^2-1)/(x^2-(y^2-1)) = 8
{b == 1, x == 3, y == 3}
etc.
3. It is easily shown that if b is square, then k must be one less than a square, say s^2-1:
k(x^2-y^2+1)=y^2-1
(k+1)(x^2-y^2+1)=y^2-1+x^2-y^2+1 adding to both sides
(k+1)(x^2-y^2+1)=x^2
So if b=(x^2-y^2+1) is a square, then k must be one less than a square, i.e. s^2-1.
4. b not only may, but must be square. Let k=s^2-1:
Then:
x = (s*sqrt(y^2-1))/sqrt(s^2-1)
y = sqrt(s^2 (x^2+1)-x^2)/s
s = x/sqrt(x^2-y^2+1)
But b = (x^2-y^2+1), so b must be square to clear the square root under x.
5. s must be odd. The respective recurrence relations of x,y,b are:
x = s*((s+Sqrt(s^2-1))^(n)-(s-Sqrt(s^2-1))^(n))/(2*Sqrt(s^2-1))
y = 1/2 sqrt((s-sqrt(s^2-1))^(2n+1)+(sqrt(s^2-1)+s)^(2n+1)+2)
b=(((s+Sqrt(s^2-1))^(n)-(s-Sqrt(s^2-1))^(n))/(2*Sqrt(s^2-1)))^2
for all n,s.
If s is even, then x must be, because the equality for x would then be multiplied by an even number. So s must be odd.
6. If s is odd, then y is also:
(s^2-1)(x^2-y^2+1)=y^2-1
s^2(x^2-y^2+1)=y^2-1+x^2-y^2+1 adding to both sides
s^2(x^2-y^2+1)=x^2
(s^2-1)(x^2-y^2+1)+1=y^2
If s^2 is odd, then (s^2-1) is even, so y is always odd.
7. Since b = x^2-y^2+1, with odd y, the parity of b is the same as that of x.
8. We need now only consider the corresponding Chebyshev polynomials for ascending powers of n. Small values of (x) = x suffice:
n=1 1
n=2 2(x)
n=3 4(x)^2-8(x)+3
n=4 4(2(x)^3+6(x)^2+5(x)+1)
n=5 16(x)^4-64(x)^3+84(x)^2-40(x)+5
etc.
If n is even, x is even, so only odd n will produce solutions with both x and y odd.
The qualifying values of x,y,b, are then as follows:
x = s*((s+Sqrt(s^2-1))^(2n-1)-(s-Sqrt(s^2-1))^(2n-1))/(2*Sqrt(s^2-1))
y = 1/2 sqrt((s+Sqrt(s^2-1) )^(4n-2)+(s-Sqrt(s^2-1))^(4n-2)+2)
b = (((s+Sqrt(s^2-1))^(2n-1)-(s-Sqrt(s^2-1))^(2n-1))/(2*Sqrt(s^2-1)))^2
for all n,s.
Edited on October 5, 2016, 4:26 am
|
|
Posted by broll
on 2016-10-05 04:13:49 |
Possible solution
