Suppose we want to solve the following Diophantine equation
$$
3a^2 + b^2 = c^2
$$
for all triples of nonnegative integers $(a, b, c)$ where $\gcd(a, b, c) = 1$ and $a \le b$. Given the solution $(0, 1, 1)$, one can use the method mentioned on Wolfram Mathworld to generate many more solutions using pairs of integers $(u, v)$:
\begin{align}
A &= 2uv \\
B &= 3u^2 – v^2 \\
C &= 3u^2 + v^2 \\
\end{align}
These $A, B, C$ are not necessarily coprime or nonnegative, so we produce $(a, b, c)$ as follows:
$$
g = \gcd(A, B, C) \qquad
(a,b,c) = \left(\frac{|A|}g, \frac{|B|}g, \frac{|C|}g\right)
$$
That much is easy. However, we can go further.
After performing brute force calculations, I am convinced that there is in fact a one-to-one correspondence between:
- the desired solutions $(a, b, c)$ as previously described,
- and coprime pairs $(u, v)$ of nonnegative integers, $0 <= u <= v$.
It is easy to prove that every $(u, v)$ admits an $(a, b, c)$, as I have already demonstrated above.
But how can one prove the converse? Or more effectively: Given $(a, b, c)$, how do we solve for $(u, v)$?
Best Answer
As coming from an identity, in this case $$3(2uv)^2+(3u^2-v^2)^2=(3u^2+v^2)^2$$ the given parameterization gives all the integer solutions of the equation $3a^3+b^2=c^2$. If $(a,b,c)$ is an integer solution the system involving the corresponding parameters $(u.v)$ easily gives the solution $$u=\sqrt{\frac{c+b}{6}}\\v=\sqrt{\frac{c-b}{2}}$$ so the parameters are not necessarily rational which is not matter of rejection. Above it has been say that the identity gives ALL the solutions but it is not said that the parameters should be necessarily rational. For example $(u,v)=(2\sqrt3,3\sqrt3)$ gives $(A,B,C)=(9\cdot4,9\cdot1,9\cdot7)$ corresponding to the solution in coprimes $(a,b,c)=(4,1,7)$