Nope. Here's how I derived the formula. Let -8 + 36*a - 27*a^2 = b^2. So, 3 = (2+b)/(3a-2) * (2-b)/(3a-2). Let (2+b)/(3a-2) = k, a rational number. So, (2-b)/(3a-2) = 3/k. Now we can easily solve for a, b, and we have a = 2/3 + (4/3)*k/(k^2+3), b = 2*(k^2-3)/(k^2+3). Similarly, this technic can be used to find a parametric solution for Pythagorean triples. Warut On Sun, Aug 2, 2015 at 10:51 PM, 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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