There's a general method for quadrics: if f(x_1,...,x_n) is a quadric (i.e. degree =2) with rational coefficients, and you know one solution (say 0, by translation), then you can get all other rational solutions. Suppose that H is a rational hyperplane not passing through 0. If a is a non-zero solution, then the line passing through 0 and a intersects H in exactly one point, whose coordinates must be rational (it might be a point at "infinity" if the line is parallel to H). Conversely, for every rational point b on H, the line through b and 0 must given you a non-zero point on the quadric, since plugging in the equation of the line x = bt, gives you a quadratic in t, and 0 is one root, so the other root is - the coefficient of t. Victor On Sun, Aug 2, 2015 at 11:51 AM, rwg <rwg@sdf.org> wrote:
Neat! You got that somehow from
In[407]:= Simplify[(2 t/(1 + t^2))^2 + ((1 - t^2)/(1 + t^2))^2]
Out[407]= 1 ? --rwg
On 2015-08-02 05:56, Warut Roonguthai wrote:
I didn't follow math-fun for a long time (i.e., don't know where this equation came from), but the new subject caught my eye.
Using an elementary method, I found that -8 + 36*a - 27*a^2 is a rational square iff
a = 2/3 + (4/3)*k/(k^2+3) for some rational k.
On Sat, Aug 1, 2015 at 11:53 PM, Bill Gosper <billgosper@gmail.com> wrote:
The data are intriguing. Something about 3 and powers of primes 1 mod 6.
What subject is this? BQFs? --rwg
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