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Trying to solve a problem I meet on the way a question, easy maybe, that I can't solve for now. The Pell-Fermat equation $$x^2-dy^2=1 \hspace{2cm}(*)$$ where the integer $d$ is square-free, has in general infinitely many integer solutions for which the fundamental unit of $\mathbb Q(\sqrt d)$ is the key. In fact, all the integer solutions $(x_n,y_n)$ of $(*)$ are given by$$x_n+y_n\sqrt d=(x_1+y_1\sqrt d)^n;\space n\ge 1$$ where $ x_1+y_1\sqrt d$ is the mentioned fundamental unit.

The punctual question is, are there non-integer rational solutions $(x,y)$ with $xy\ne0$ for $x^2-3y^2=1?$

(All the integer solutions are given by $x_n+y_n\sqrt3=(2+\sqrt3)^n$)

Piquito
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  • All the integer solutions are given by $x_n + y_n\sqrt{3} = \pm(2+\sqrt{3})^n$ for $n$ in $\mathbf Z$. – KCd Sep 07 '22 at 17:24

1 Answers1

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This sort of thing is parametrized by a finite number of recipes of the Pythagorean triple type. It is not necessarily parametrized by just one recipe. A reference is Mordell, Diophantine Equations, page 47, Theorem 4.

Multiplying through be common denominator $w$ leads to integers $u,v,w$ with $$ u^2 - 3 v^2 = w^2 $$ or $$ u^2 = 3 v^2 + w^2 \; . \;$$

If, in turn, we demand $u,v,w$ coprime, ... with coprime integers $m,n:$

When $u$ is even we get all solutions by $$ u = 2 m^2 + 2mn + 2 n^2 \;, \; \; v = m^2 + 2mn \; , \; \; w = -m^2 + 2mn + 2 n^2 \; . \; $$

When $u$ is odd we get all solutions by $$ u = 3 m^2 + n^2 \;, \; \; v = 2mn \; , \; \; w = 3m^2 - n^2 \; . \; $$

In the first one we need $m$ odd, not sure about conditions $\pmod 3$

In the second one we need $m+n$ odd and $ n \neq 0\pmod 3$

Will Jagy
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  • Thanks you. This question can be related to an elementary answer to the problem you have solved with " heavy artillery" (in which not- mentioned elliptic curves are involved) in https://math.stackexchange.com/questions/4525135/a2-b2-c2-ab-6d2-a-b-1-has-no-solutions-a-b-c-d-in-mat . In fact we have the equation $$(a+b)^2-3(2d)^2=c^2$$ for which the integer solutions of $x^2-3y^2=1$ are not good. – Piquito Sep 06 '22 at 19:31
  • @Piquito I found a good answer to that one at https://math.stackexchange.com/questions/4218013/why-p-equiv-3-mod-8-is-not-a-congruent-number/4222489#4222489 by Dan Fulea – Will Jagy Sep 06 '22 at 19:36
  • Thank you very much for the interesting link. The “objection” to the post of a respectable MSE member made me smile: “Are not congruent to what?” – Piquito Sep 06 '22 at 20:00