Difference between revisions of "Pell equation"
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<math>x_{m}+\sqrt{D}y_{m} < p+\sqrt{D}q < x_{m+1}+\sqrt{D}y_{m+1}</math> | <math>x_{m}+\sqrt{D}y_{m} < p+\sqrt{D}q < x_{m+1}+\sqrt{D}y_{m+1}</math> | ||
− | Next, multiply the inequality by <math> x_{m}-\sqrt{D}y_{m} to obtain: | + | Next, multiply the inequality by <math> x_{m}-\sqrt{D}y_{m}</math> to obtain: |
− | 1 < (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m}) = (px_{m}-Dqy_{m})+\sqrt{D}(qx_{m}-py_{m})< x_{0}+y_{0}\sqrt{D}< | + | 1 < (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m}) = (px_{m}-Dqy_{m})+\sqrt{D}(qx_{m}-py_{m})< x_{0}+y_{0}\sqrt{D}<math>. |
However, it can be seen that | However, it can be seen that | ||
− | <math>(px_{m}-Dqy_{m})^2-D(qx_{m}-py_{m})^2 = (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m})(p-\sqrt{D}q)(x_{m}+\sqrt{D}y_{m}) = 1< | + | </math>(px_{m}-Dqy_{m})^2-D(qx_{m}-py_{m})^2 = (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m})(p-\sqrt{D}q)(x_{m}+\sqrt{D}y_{m}) = 1<math> |
− | Meaning <math>(px_{m}-Dqy_{m}, qx_{m}-py_{m})< | + | Meaning </math>(px_{m}-Dqy_{m}, qx_{m}-py_{m})<math> is a solution smaller than the minimal solution which is a contradiction. |
− | Therefore, such <math>(p,q) | + | Therefore, such </math>(p,q)$ cannot exist and so the method of composition generates every possible solution to Pell's equation. |
Q.E.D. | Q.E.D. |
Revision as of 12:17, 23 June 2018
A Pell equation is a type of diophantine equation in the form for a natural number . Generally, is taken to be square-free, since otherwise we can "absorb" the largest square factor into by setting .
Note that if is a perfect square, then this problem can be solved using difference of squares. We would have , from which we can use casework to quickly determine the solutions.
Alternatively, if D is a nonsquare then there are infinitely many distinct solutions to the pell equation. To prove this it must first be shown that there is a single solution to the pell equation.
Claim: If D is a positive integer that is not a perfect square, then the equation has a solution in positive integers.
Proof: Let be an integer greater than 1. We will show that there exists integers and such that with . Consider the sequence . By the pigeon hole principle it can be seen that there exists i, j, and p such that i < j, and
.
So we now have
.
We can now create a sequence of such that and which implies r and s. However we can see by the pigeon hole principle that there is another infinite sequence which will be denoted by such that . Once again, from the pigeon hole principle we can see that there exist integers f and g such that mod H, mod H, and . Define and notice that . Also note that mod H which means that Y = 0 mod H also. We can now see that is a nontrivial solution to pell's equation.
Family of solutions
Let be the minimal solution to the equation . Note that if are solutions to this equation then which means is another solution. From this we can guess that is obtained from . This does indeed generate all the solutions to this equation. Assume there was another solution . If is non-minimal, then there exists some integer such that
Next, multiply the inequality by to obtain:
1 < (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m}) = (px_{m}-Dqy_{m})+\sqrt{D}(qx_{m}-py_{m})< x_{0}+y_{0}\sqrt{D}$.
However, it can be seen that$ (Error compiling LaTeX. Unknown error_msg)(px_{m}-Dqy_{m})^2-D(qx_{m}-py_{m})^2 = (p+\sqrt{D}q)(x_{m}-\sqrt{D}y_{m})(p-\sqrt{D}q)(x_{m}+\sqrt{D}y_{m}) = 1(px_{m}-Dqy_{m}, qx_{m}-py_{m})$is a solution smaller than the minimal solution which is a contradiction.
Therefore, such$ (Error compiling LaTeX. Unknown error_msg)(p,q)$ cannot exist and so the method of composition generates every possible solution to Pell's equation.
Q.E.D.
Continued fractions
The solutions to the Pell equation when is not a perfect square are connected to the continued fraction expansion of . If is the period of the continued fraction and is the th convergent, all solutions to the Pell equation are in the form for positive integer .
Generalization
A Pell-like equation is a diophantine equation of the form , where is a natural number and is an integer.
Introductory Problems
Show that if and are the solutions to the equation , then .