A study of some Diophantine equations (Q2741814)

The author proves the following theorems. Theorem 1: If \(D\equiv 2,5\bmod 8\), \(D>0\), and the Pell equation \(x^2-Dy^2=-1\) has an integral solution \(x,y\), then the Diophantine equation \(x^2-Dy^{2n}=1\) \((n>2)\) has no integral solution \(x,y\). Theorem 2: The Diophantine equation \(4x^4-Dy^2=1\) (\(D>0\), \(D\) not a square) has a solution in positive integers \(x,y\) if and only if there are integers \(x_1,y_1\) such that \(2x^2_1+y_1\sqrt D=\epsilon\), where \(\epsilon\) is the fundamental solution of the Pell equation \(x^2-Dy^2=1\). Theorem 3: If the Diophantine equation \(x^4-Dy^2=-1\) (\(D>0\), \(D\) not a square) has a solution in positive integers \(x,y\), then \(x^2+y\sqrt D=\eta^d\), where \(\eta=u_0+v_0\sqrt D\) is the fundamental solution of the Pell equation \(x^2-Dy^2=-1\), \(u_0=du_1^2\), \(d\) is a square-free integer. Let \(p\) be an odd prime \(>3\). In order to prove that the equation \(x^2-1=y^p\) has no integer solutions with \(x>1\), \textit{C. Ko} calculated the Jacobi symbol \((Q_p(y)/Q_q(y))\), where \(Q_n(y)=(y^n+1)/(y+1)\), \(2\nmid n\), and \(q\) is an odd prime with \(q\not= p\) [Sci. Sin. (Notes) 14, 457-460 (1964)]. The proofs of Theorems 2 and 3 depend upon an application of Ko's method.