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 \(k\).
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 \(x=2m+1\), \(y=2n\).
Imagine a \(2\times 2\) “Lights Out" board. If every light is off at the start, how can we turn on just one of the squares? Can you notice something about the order in which we press squares?
Suppose we have a \(2\times 2\) board where all the lights start being turned off, how can we turn on the top two lights?
A \(2\times 2\) "Lights Out" board starts with all the light being turned off. How can you turn on the top-left and bottom-right squares at the same time?
Now let’s imagine a \(3\times 3\) board, how can you turn on just the middle light?