If we know the minimal solution or any specific solution of Pell's equation $x^2-ny^2=1$ , is there is any general formula to write all solution of Pell's equation?
General solution of Pell's equation
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On
$$a^2-db^2=\pm1\iff(a+b\sqrt d)(a-b\sqrt d)=\pm1\iff(a\pm b\sqrt d)\text{ are units in } \mathbb Q(\sqrt d)$$Hence the elementary theory of algebraic numbers gives all the solutions $(a_n,b_n)$ are given by the equations $$(a_n+b_n\sqrt d)=(a_0+b_0\sqrt d)^n$$ and $$(a_n+b_n\sqrt d)=2^{1-n}(a_0+b_0\sqrt d)^n$$ according if $d\equiv 2,3\pmod 4$ and $d\equiv 1\pmod 4$ respectively and where $a_0+b_0\sqrt d$ is the fondamental unit of the ring of integers of the quadratic field $\mathbb Q(\sqrt d)$.
On
Another way to understand this is that the solutions $(a,b)$ to $a^2-db^2=1$ are a subgroup of the group of units in the ring $\mathbb Z[\sqrt{d}]$, because the property $a^2-db^2=1$ is the assertion that the Galois norm $N(a+b\sqrt{d})$ of $a+b\sqrt{d}$ is $1$. The Galois norm is multiplicative, meaning that $N(\alpha\cdot \beta)=N(\alpha)\cdot N(\beta)$. It is a very special case of Dirichlet's Units Theorem that the group of such things, disregarding $\pm 1$, is a free abelian group on one generator. That is, everything of norm $1$ in that ring is $\pm(a_o+b_o\sqrt{d})^N$ for one fixed $a_o+b_o\sqrt{d}$ and integer exponent $N$. (This $a_o+b_o\sqrt{d}$ is called the "fundamental unit".)
Then the $a,b$ of solutions to Pell's equation are the rational part and coefficient of the irrational part in that power of the fundamental unit.
On
Ummm; if you have $u,v$ the smallest positive (nonzero) solution to $u^2 - n v^2 = 1,$ then as in the answer by Stefan4024, all solutions are obtained from the $(1,0)$ solution by the mapping $$ (x,y) \mapsto (ux + nvy, \; v x + u y). $$ The Cayley-Hamiltion Theorem says that we can write separate linear recurrences for $x,y.$ That is $$ x_0 = 1, x_1 = u, x_2 = 2 u^2 - 1,$$ and $$ x_{n+2} = 2 u x_{n+1} - x_n. $$
$$ y_0 = 0, y_1 = v, y_2 = 2 u v,$$ and $$ y_{n+2} = 2 u y_{n+1} - y_n. $$
The full proof that this gives all solutions is rather long. Maybe I should just draw attention to alternatives. All solutions to $x^2 - 5 y^2 = 6061$ make up a rather more complicated set; in the sense discussed above, there are eight orbits, not just one.
jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
Automorphism matrix:
9 20
4 9
Automorphism backwards:
9 -20
-4 9
9^2 - 5 4^2 = 1
x^2 - 5 y^2 = 6061
Sun Jul 3 14:58:33 PDT 2016
x: 79 y: 6 ratio: 13.1667 SEED KEEP +-
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x: 191 y: 78 ratio: 2.44872 SEED BACK ONE STEP 159 , -62
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Sun Jul 3 14:59:13 PDT 2016
x^2 - 5 y^2 = 6061
jagy@phobeusjunior:~$
jagy@phobeusjunior:~$
It's well know that the solution of the Pell's equation satisfy the reccurence relation:
$$x_{k+1} = x_1x_k + y_1ny_k$$ $$y_{k+1} = y_1x_k + x_1y_k$$
This was found by Brahmagupta and it holds because:
$$1 = (x_1 - ny_1)(x_k - ny_k) = (x_1x_k + y_1ny_k) - n(y_1x_k + x_1y_k)$$