Euler called  x² + m y²  and  x² + n y²  concordant forms  if they can be both made integer squares with integers x and y with xy ≠ 0 (assuming m and n are also integers).

Dickson's History of the Theory of Numbers (vol 2) devotes 6 pages in Chapter 16 to this topic.

Euler, himself, considered the problem with m=1, which is what we consider in this simple essay.

We thus look for non-zero integer solutions of

x2 + y2  =  a2                        x2 + N y2  =  b2

We can assume that x and y have no common factors, so that standard Pythagorean triple results imply that

1. x = p² - q²  and  y = 2 p q

 
2. x = 2 p q  and  y = p² - q²

which, if we define  y = b / q²  and  x = p / q, gives

y2  =  x4 + ( 4 N - 2 ) x2 + 1

This has the obvious rational point  x = 0,  y = 1, and so is birationally equivalent to an elliptic curve. Using the standard algebra, gives the family of curves

EN: v2  =  u3 + ( N + 1 ) u2 + N u  =  u ( u + 1 ) ( u + N )

with the transformation   p / q = v / ( u + N ).

The second x-y parameterisation leads to

b2  =  N p4 + 2 ( 2 - N ) p2 q2 + N q4

which, with (as before)   y = b / q²   and   x = p / q, gives

y2  =  N x4 + ( 4 - 2 N ) x2 + N

Since x = 1,  y = 2 lies on this curve, it is birationally equivalent to an elliptic curve. We get the same curve as before, though with the transformation

p  q   =   - u + v + N  u - v + N

As an example, if  N = 59, the curve E₅₉ : v² = u³ + 60 u² + 59 u has a point
u = -219539 / 212521,  v = 137769720 / 97972181.

From the first transformation we get   p / q = 671 / 27660 which gives
x = 764625359 and y = 37119720. These satisfy

x2 + y2  =  7655258412

and

x2 + 59 y2  =  8160554412

The second transformation gives  p / q = -28331 / 26989 giving x and y values which are exactly twice those from the first transformation.

The curves E_N have 3 obvious finite points of order 2,  u = 0,  u = -1 and u = -N, none of which lead to non-zero solutions. If N = k², there are 4 points of order 4, namely  u = k,  v = ± k ( k + 1 ) and u = -k,  v = ± k ( k - 1 ), again which do not give non-trivial solutions. There is a paper by Ono [ono], which proves these are the only torsion points.

Thus, to find solutions, we need the rank of E_N to be at least 1. We considered N between 2 and 9999, and the Birch and Swinnerton-Dyer conjecture was used to estimate the rank. The following table gives the distribution of the rank in this range.
 

The computed results for these values have been split into 4 file:

(a)  N in [ 2 , 2499 ]
(b)  N in [ 2500 , 4999 ]
(c)  N in [ 5000 , 7499 ]
(d)  N in [ 7500 , 9999 ]
 

As of 10 January 2005, all values in 2-9999 have been solved. Many thanks are due to Randall Rathbun who finished off most of the remaining difficult problems during Xmas and New Year.
 

If I get the energy, I might extend the values further. Randall tells me he is looking at the case of N negative, so I'll leave that alone.

At the beginning of July 2005, Randall sent me a file with solutions for N in [-9999,-1]. The 39 unsolved values all have estimated heights of over 200 so will be quite difficult to solve.