This is a somewhat different example than most of the others, in that we will see that solutions exist for all integer N > 1. We will use elliptic curves to derive some parametric solutions.

Suppose that

N2 x2  +  N x y  +  y2  =   b2

and define  w = b / y  and  z = x / y, so that

w2  =  N2 z2 + N z + 1

This has the obvious point  z = 0, w  = 1. The line w = 1 + m z meets the quadratic at z = 0 and
z = ( N - 2  m ) /  ( m² - N² ).

Let m = p / q,  giving the parameterisation x = q ( 2 p - N q ) and y = N² q² - p².

We now substitute this into the first quadratic

a2  =  x2 + N x y +N2 y2

giving

a2 = N2p4 - 2Np3q - (2N4-N2-4)p2q2 + (2N3-4N)pq3 + N2(N4-N2+1)q4

Define   k = N a / q²   and   j = N p / q ,  giving

k2 = j4 - 2 j3 - ( 2 N4 - N2 - 4 ) j2 + 2 N2 ( N2 - 2 ) j + N4 ( N4 - N2 + 1 )

which has the rational solution  j = k = N² , and so is birationally equivalent to an elliptic curve.

After some standard manipulation, we find this curve to be

EN:  v2 = u3 + 2 ( 2 N4 - N2 + 2 ) u2 + 9 N4 u

with the (u,v) ® (j,k) transformation leading to the relation
 

p q =  3 N2 + u + v 2 N ( u + 3 )

There is only one finite point of order 2, namely (0,0).

There are two points of order 4, (-3N² , ±6 N² (N² - 1 ) ).

Because the first coordinate of this point is < 0, there cannot be points of order 8. This suggests that there are only 4 torsion points (including the pt at infinity), and the torsion subgroup is isomorphic to Z4. I would like a simple proof of this.

If you look at the p/q relation, the denominator becomes zero if  u = -3. I have always found that such observations lead to interesting points, and we find
u = -3 ,  v = ±3 ( N² - 1 ).

This point is not a point of inflexion if N > 1. We thus suspect that this is a point of infinite order.

If we double the point ( -3, 3 ( N² - 1 ) ) we have u = 9 ( N² + 1 )² / 4. If N is even, this value will be a non-integer rational and so must be a point of infinite order. For N odd, the value of u will be an integer, and so possibly a torsion point.

If we add (0,0) to this point we get u = 4 N⁴ / ( N² + 1 )²  < 4,  and since u = 1, 2, 3 don't give rational points, we must have u non-integer and hence a point of infinite order.

Thus the points  (-3, ±3(N²-1) ) are of infinite order if N > 1 and so the curves E_N have rank ³ 1 for N > 1.

Since this point leads to an undefined relation for p/q, we use the double point with
v = -9 (N²+1) (5N⁴+6N²+5)/8, which leads to
p = -15N⁶+27N⁴+13N²+9  and  q = 4N(3N⁴+6N²+7).

Using the parameterisation leads to

x = -24 N ( N2 + 1 ) ( 3N4 + 6N2 + 7 ) ( 7N4 + 6N2 + 3 )

and

y = -3 ( N+1)2 (N-1)2 (N2+3) (3N2+1) (3N2+2N+3) (3N2-2N+3)

From these values we find

a = 3 N ( 27N12 + 162N10 + 673N8 + 1164N6 + 1197N4 + 690N2 + 183 )

and

b = 3 ( 183N12 + 690N10 + 1197N8 + 1164N6 + 673N4 + 162N2 + 27 )

Thus we have used elliptic curves to show that solutions exist for all N > 1, and to provide one parametric solution. Other solutions can be found by using other points on the curves generated from the pt of infinite order combined with the torsion points.

What I would like to know is:

• is this result obvious?
• can it be shown without the use of elliptic curves?
• are there other ways of finding parametric solutions?