NT-B2-M08-P020 The Equation \(x^2+y^2+3=4xy\)

Find the base solution for \(x=1\), then apply Vieta jumping.

Let \(x\le y\). For \(x=1\), \(y^2-4y+4=0\), so \(y=2\). For \(x=2\), \(y^2-8y+7=0\), so \(y=1\) or \(7\); this matches neighbouring terms of the future sequence. For \(x=3\), the discriminant \(144-48=96\) is not a square. For \(x\ge 4\), the second root is \(y'=4x-y=\frac{x^2+3}{y}\), positive and integral. Since \(y\ge x\), \(y'\le x+\frac{3}{x}

Reversing the jump: if \((a,b)\) is a solution, then \((b,4b-a)\) is also a solution. Starting from \((1,2)\), define \(a_0=1\), \(a_1=2\), \(a_{n+1}=4a_n-a_{n-1}\). All solutions are \((a_n,a_{n+1})\) and swaps, \(n\ge 0\).

Final problem of the module: similar to the central one, but with different base cases.