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 . Then there exists some m such that
However, it can be seen that
Meaning is a solution smaller than the minimal solution which is a contradiction. This article is a stub. Help us out by.
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 .
A Pell-like equation is a diophantine equation of the form , where is a natural number and is an integer.
If and are the solutions to the equation , then .