Find the mistake in the sequence of equalities: \(-1=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1\).
Two players are playing a game. The first player is thinking of a finite sequence of positive integers \(a_1\), \(a_2\), ..., \(a_n\). The second player can try to find the first player’s sequence by naming their own sequence \(b_1\), \(b_2\), ..., \(b_n\). After this, the first player will give the result \(a_1b_1 + a_2b_2 + ...+a_nb_n\). Then the second player can say another sequence \(c_1\), \(c_2\), ..., \(c_n\) to get another answer \(a_1c_1+ a_2c_2 + ... +a_nc_n\) from the first player. Find the smallest number of sequences the second player has to name to find out the sequence \(a_1\), \(a_2\), ..., \(a_n\).
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 \(y\).
\(x^2 - 7y^2 = 1\);
\(x^2 - 7y^2 = 29\).