Quadratic integer equation

Question

I am trying to find integer solutions to $\space 2x^2-y^2=1\space $ other than $\space (1,1).\quad$ Just one is sufficient, and I know (by brute force) that $\space (5,7)\space $ is one. How do I show my work?

Answer 1

A "Y-solution" and a spreadsheet show a lot. It is semi-brute force but not hard. Make an $\space x$-column with sequential numbers where $\quad n\le x \le 1000+x.\quad$ in the next column put the formula from the RHS of "y=". In the next column over enter $\quad\text{IF(Y-value=INT(Y-value),"***","").}\quad$ In the top row of this column use "COUNTIF()" to count the number of asterisks down to the thousandth row. If the count is greater than zero, search the rows below for integers. Change the first number in the $\space x$-column up by $1000$ to continue counting 1000 at a time.

\begin{align*} 2x^2-y^2&=1\\ \implies 2x^2 y^2 - 1 &= y^2\\ \implies y=\pm&\sqrt{2 x^2-1}\\ \implies (x,y)\in\pm& \big\{(1,1),(5,7),(29,41),(169,239),(985,1393),\cdots\big\} \end{align*}

$\textbf{Update: }\space $ I just noticed a pattern in this series. All $\space x$-values can be generated by the following formula which generates every other Pell Number which means that the $\space n^{th}\space x\space$ can be generated directly.

$$ x= \frac{(1 + \sqrt{2})^{(2n-1)} - (1 - \sqrt{2})^{(2n-1)}}{2\sqrt{2}} $$