The Diophantine equation that I have to solve is: $$343x^2-27y^2=1$$ This question has already been posted by other user but it has not received an answer. I proved to solve it.

This is my attempt:

substituting for $x^2=u$ and $y^2=v$ the equation becomes a Diophantine linear equation:$$343u-27v=1$$ using Euclid's algorithm to solve the equation, the solutions are: $$ \left\{ \begin{array}{ll} u=10+27k \\ v=127+343k\end{array} \right. $$ substituting these solutions in first equation we obtain: $$y^2=127+343\cdot (\frac {x^2-10}{27})$$ $x^2-10$ has to be a multiple of $27$ therefore $$x^2=c27+10 \tag{1}$$ (with $c$ integer)

$(1)$ becomes $$(x-1)(x+1)=9\cdot (3c+1)$$ obtaining two systems: $$ \left\{ \begin{array}{ll} x+1=n9 \\ x-1=\frac {3c+1}{n}\end{array} \right. $$ and $$\left\{ \begin{array}{ll} x-1=n9 \\ x+1=\frac {3c+1}{n}\end{array} \right. $$ The solution of the first is $c=3p+1$ (with $p$ integer): indeed if $x+1\equiv 0\pmod 3$ $c$ has to be $c\equiv 1\pmod 3$ because $c(x-1)-1$ has to be a multiple of $3$. The solution of the second system is $c=3p+2$: indeed if $x-1\equiv 0\pmod 3$ $c\equiv 2\pmod 3$. How can I can continue and are there other solutions?

  • 1
    $\begingroup$ Please link to the prior question. $\endgroup$ – Bill Dubuque May 17 '15 at 17:21
  • $\begingroup$ The question has been closed $\endgroup$ – Domenico Vuono May 17 '15 at 17:22
  • $\begingroup$ But you can still link to it. Maybe it will get reopened. $\endgroup$ – Bill Dubuque May 17 '15 at 17:23
  • $\begingroup$ @BillDubuque math.stackexchange.com/q/1282128/227073 $\endgroup$ – Domenico Vuono May 17 '15 at 17:27
  • $\begingroup$ 376766, 1342879 ? $\endgroup$ – Yimin May 17 '15 at 17:28

For generalization of the Pell equation question arises constantly. For example in this thread.

Whenever Pell's equation proof is solvable, it has infinitely many solutions

So writes the Pell equation in General form.


If we know any solution of this equation. $( p ; s)$

If we use any solutions of the following equation Pell.


Then the following solution of the desired equation can be found by the formula.





$⇒343x^2≡1 mod 27$

$343≡19 mod 27$

we can find that if $x=8$ we get:

$343 . 8^2 ≡ 1 mod 27$

We can see that x belong to a set its members make an arithmetic progression and have following specification:

$s . 19 ≡ 1 mod 27$; $s = [10, 37, 64, . . a+27k]$

where $a=10$; and $d=27$ is common ratio.

With $k=5257504398$ we get $x=376766$

Also $y^2=343 k$ that gives $y=1342879$

Therefore the parametric solution is:


$y=\sqrt{343 k}$

This equation probably has more solutions.


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