Suppose that \(x_1+y_1\sqrt{d}\) and \(x_2+y_2\sqrt{d}\) give solutions to Pell’s equation \(x^2-dy^2=1\) and \(x_1,x_2,y_1,y_2\geq 0\). Show that the following are equivalent:
\(x_1+y_1\sqrt{d} < x_2+y_2\sqrt{d}\),
\(x_1<x_2\) and \(y_1<y_2\),
\(x_1<x_2\) or \(y_1<y_2\).
If Pell’s equation \(x^2-dy^2 = 1\) has a nontrivial solution \((x_1,y_1)\), show that it has infinitely many distinct solutions.
Show that there are infinitely many triples of consecutive integers, each of which is a sum of the square of two integers.
Suppose that Pell’s equation \(x^2-dy^2=1\) has a solution \((x_1,y_1)\) where \(x_1,y_1\) are positive and \(y_1\) is minimal among all solutions with positive \(x,y\). Show that if \(x+y\sqrt{d}\) gives a solution to \(x^2-dy^2=1\), then \(x+y\sqrt{d}=\pm(x_1+y_1\sqrt{d})^k\) for some integer \(k\).
The original “Lights Out” game works like this: a light pattern is shown on the board, and your task is to turn all the lights off. A light pattern is called solvable if you can complete the game starting from that pattern. Ziheng and Jan are playing on an \(n\times n\) board, and they notice that some patterns are unsolvable. Can you find a rule to decide when a pattern is not solvable?
Suppose that \(x_1+y_1\sqrt{d}\) gives a solution to Pell’s equation \(x^2-dy^2=1\). Define a sequence \(x_n+y_n\sqrt{d} = (x_1+y_1\sqrt{d})^n\). Show that we have the recurrence relations \(x_{n+2} = 2x_1x_{n+1}-x_n\) and \(y_{n+2} = 2x_1y_{n+1}-y_n\).
Prove that the only solution to \(5^a-3^b=2\) with \(a,b\) being positive integers is \(a=b=1\).
Show that Pell’s equation \(x^2-dy^2=1\) has a nontrivial solution.
For the following equations, find the integer solution \((x,y)\) with the smallest possible absolute value of \(y\).
\(x^2 - 7y^2 = 1\);
\(x^2 - 7y^2 = 29\).
Find the integer solution \((x,y)\) with the smallest possible absolute value of \(y\). \(x^2 - 2y^2 = 1\);
This equation helps to find all the square-triangular numbers, namely all the numbers that are perfect squares and can be represented as the sum \(1+2+3+...m\) for some \(m\). Finding such a number is equivalent to finding a solution to the equation: \(2n^2 = m(m+1)\). Or finding a solution to the Pell’s equation \(x^2-2y^2 = 1\) for \(x=2m+1\), \(y=2n\).