The equation has the obvious solutions
(1) x = a , y = 0 , z = a²
(2) x = 0 , y = b , z = b²
for any m, so we restrict attention to solutions with non-zero values.

The equation of the title can be put into the form

t2 = w4 + m w2 + 1

by using the substitutions  t = z / y²  and  w = x / y.

This has the obvious rational point  w = 0 , t = 1 , and so is birationally equivalent to an elliptic curve.

Using standard methods , we find the curve

Em  :  v2 = u 3 - 2 m u2 + ( m2 - 4 ) u

with the transformation formula  w = x / y  = v / ( 2 u ).

This is not the form given by Bremner and Jones who discuss

s2  =  r3 + m r2 + r

which is the 2-isogenous curve to E_m. We find that E_m has rational points of smaller height, on average.

As an example,consider the example discussed by Bremner and Jones, where m = -195.

The curve is  v² = u³ + 390u² + 38021u with the Birch and Swinnerton-Dyer conjecture predicting rank 1 with height 14.47. The methods we use to find points give

u = 36175850761057 / 76494283776
and
v = 307299844206714529135 / 21156483029630976

The transformation  x / y = v / (2u) gives

x=3677683140865
and
y=239483286336
giving
z=5628067524800453986859521, as given by Bremner and Jones.

It is interesting to note that the 2-isogeny curve of Bremner and Jones  s² = r³ - 195r² + r has a point with height 28.94, which is much more difficult to find.

The cubic  u³ - 2 m u² + (m²-4) u  =  u (u-m+2) (u -m-2) has clearly 3 rational roots, so there are 3 points of order 2,  (0,0), (m+2,0), (m -2,0), all of which give the trivial solutions to the original quartic. In general, these are the only torsion points, except when m=n² -2. In this case there are 4 points of order 4,  ( n(n+2) , ±2n(n+2) ) and  ( n(n -2) , ±2n(n - 2) ) , which correspond to the obvious solutions x = ±1 , y = ±1 , z = ±n  to the original problem.

Is there a simple proof that these are the only torsion points?

If there is, this implies that we need rank(E _m) > 0. We used the Birch and Swinnerton-Dyer conjecture over the range |m| < 10000 to estimate the rank. A summary table of the results is given below.
 
 
 

Can anybody expain the different distributions for positive and negative m?

The solutions, so far found, to the original quartic can be found in the following files:

(a) for m in  [1,1999], COMPLETE
(b) for m in  [2000,3999], COMPLETE
(c) for m in  [4000,5999], COMPLETE
(d) for m in  [6000,7999], COMPLETE
(e) for m in  [8000,9999], COMPLETE
(f) for m in  [-1999,-1], COMPLETE
(g) for m in  [-3999,-2000], COMPLETE
(h) for m in  [-5999,-4000], COMPLETE
(i)  for m in  [-7999,-6000], COMPLETE
(j)  for m in  [-9999,-8000], COMPLETE

I finally finished the remaining unknowns during July 2006. I used a mixture of Heegner-pt calculations together with 4-descent which speeds up the process. I still needed to use a distributed form of the Heegner-pt summation, using 2 labs of student PCs. Thankfully, students go on holiday at this time of year!!

I have NO energy to extend this data-set at the moment.