Seven Smurfs live in seven mushroom houses. There is a tunnel between every pair of houses, so from any house you can walk to any other house. One of the Smurfs, Clumsy, starts walking from his house, but he must not use the same tunnel more than once. He keeps walking until he reaches a house where all the tunnels have already been used. Where will Clumsy’s journey end?
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 \(k\).
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 \(x=2m+1\), \(y=2n\).