Belated Birthday Problem

The Diophantine equation

$x^2 = 1 + 2y^2$

can be rewritten as

$x^2 - 2y^2 = 1.$

This is a specific instance of Pell's equation . It can be shown that if $(x, y) = (a, b)$ is the smallest solution of $x^2 - 2y^2 = 1,$ then $(m, n)$ such that $m + n\sqrt{2} = (a + b\sqrt{2})^k,$ for some positive integer $k,$ is also a solution. In fact, this process will produce all of the solutions to the equation.

We can easily find that $(a, b) = (3, 2)$ is the smallest solution to the given equation. The solution given in the problem, $(17, 12),$ can be found if we take $k = 2$ :

$(3 + 2 \sqrt{2})^2 = 3^2 + 2(3)(2\sqrt{2}) + 8 = 17 + 12\sqrt{2}.$

The next smallest solution after that occurs when $k = 3$ :

$(3 + 2\sqrt{2})^3 = (17 + 12 \sqrt{2})(3 + 2 \sqrt{2}) = 51 + 34 \sqrt{2} + 36 \sqrt{2} + 48 = 99 + 70\sqrt{2}.$

Therefore, the next smallest integer greater than 17 whose square is one more than twice another square is $x = \boxed{99}.$

This is not a solution, but a paper/article I found over on the internet. Its a brilliant article written by a mathematician, proving the general case using induction. I am posting it here for the benefit of others. http://www.fq.math.ca/Scanned/9-1/lafer.pdf