x2+Ny2 AND x2+(N+1)y2 BOTH SQUARES

At the end of the section "Two functions of one unknown made squares" in Chapter XVI of Dickson volume 2, there is the statement

M. Rignaux proved that 2y2+1 and 3y2+1 are both squares only when y=0.

This leads directly to the more general question:

For which N can we find non-zero rationals r such that Nr2+1 and (N+1)r2+1 are both squares?

Suppose
                          N r2 + 1 = s2                    ( N + 1 ) r2 + 1 = t2

and let r = y/x, s = z/x, and t = w/x.

Thus,
                    x2 + N y2 = z2                  x2 + ( N + 1 ) y2 = w2
and so y2=w2-z2.

We can assume that y and z have no common factor, so either
(a) z = p2 - q2 and y = 2 p q    or   (b) z = 2 p q  and y = p2 - q2.

Taking the first option, and substituting into   x2 = z2 - N y2   gives

                            x2  =  p4 - 2 ( 2 N + 1 ) p2 q2  +  q4

Define h=x/q2, g=p/q, so this quartic is now

                                  h2  =  g4 - 2 ( 2 N + 1 ) g2 + 1

which has the obvious rational point g=0,h=1, and is thus birationally equivalent to an elliptic curve.

Standard methods show that we can transform to the curves
             En : v2  =  u3 + ( 2 N + 1 ) u2 + ( N2 + N ) u
with the transformation  g  =  p/q  = -u/v.

If we take the alternative z = 2 p q , y = p2 - q2, we get the same curves but with p/q = 2u/(u-v).

As an example, if N=19, the curve v2=u3+39u2+380u has a rational point (304/9 , 8360/27) giving p/q=6/55. Ignoring minus signs, we then get z=2989, y=660, x=811 and w=3061.

The curves En have 3 finite points of order 2 when u=0,u=-n, and u=-(n+1). For points of order 4, we would need integer solutions to p2=N(N+1), which are impossible.

It is possible that there are points of order 3, but I can't find a simple proof that they don't exist, though computational results suggest that the torsion group is isomorphic to Z2×Z2.

If we assume that there are only 4 torsion points then none of them lead to non-trivial solutions. Thus to find a solution we need that rank( En ) > 0.

Using the Birch and Swinnerton-Dyer conjecture we estimated the ranks for N in the range 1 to 999. The following table gives the distribution of ranks.
 
 

            N    Rank = 0    Rank = 1  Rank > 1
         1 - 99           60          38           1
  100 - 199         47          49           4
  200 - 299         42          54           4
  300 - 399         49          50           1
  400 - 499         41          56           3
  500 - 599         45          51           4
  600 - 699         46          43         11
  700 - 799         53          44           3
  800 - 899         47           48           5
  900 - 999         42          54           4 

For curves with rank greater than 0, we then looked for a point of infinite order, and hence a non-trivial solution. The results can be found in the file dd2res .

Recently ( February/March 2004 ), I have returned to this problem with a new faster machine and using a combination of some of my codes and John Cremona's ratpoint program. This has allowed me to reduce the number of unsolved values of N to ONE!!!

The one remaining value is N=982 which has a generator with predicted height 85.9. Thus it will be difficult to find but any help would be appreciated.

No need - found solution with Heegner point calculation so results now complete for [1,999]. I might extend further, possibly to N negative.

File translated from TEX by TTH, version 2.86.
Latest revision: 24 April 2006.