For an odd prime $p$, consider the equation $\begin{equation} x^2-3y^2=p^2 \end{equation}.$ What are non-trivial integral solutions of this equation? Thanks in advance for providing the solutions.
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to find some ideas you must calculate some examples for odd primes – Dr. Sonnhard Graubner Aug 04 '16 at 10:58
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1Take a look here: http://math.stackexchange.com/questions/238587/solutions-to-diophantine-equation-x2-d-y2-m2 – Robert Z Aug 04 '16 at 11:07
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You can also use this: https://www.alpertron.com.ar/QUAD.HTM – Robert Z Aug 04 '16 at 11:12
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Please improve your title to describe exactly the question you are asking, as specific as possible. – Caleb Stanford Aug 04 '16 at 13:12
2 Answers
I wrote a C++ program that finds the solutions to this equation.
There are $80$ solutions for $y\le100\ ,\ p\le167$.
The first $10$ solutions are:
$$(x,y,p)=
(6,3,3),
(14,3,13),
(13,4,11),
(10,5,5),
(14,5,11),
(38,5,37),
(14,7,7),
(26,7,23),
(74,7,73),
(19,8,13)
$$
Trivial case: You can generate a solution by picking any prime number $p$ and setting $x=2p,\ y=p$.
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HINT.-The fondamental unit of the ring of integers of the field $\Bbb Q(\sqrt 3)$ it is known to be $u_0=2+\sqrt 3$ so the general solution of the Pell-Fermat equation $x^2-3y^2=1$ is given by $$a_n+b_n\sqrt 3=(2+\sqrt 3)^n$$ and an infinite set of solutions of the proposed equation are $$(x,y)=(pa_n,pb_n)$$ in which you have to discriminate when the prime $p$ is fixed or variable.
For an immediate solution (for $n=1$) you have $$(x,y)=(2p,p)$$
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