What are the nonegative integral solutions to $k^2 = 2 n^2 +1$?

Question

What are the nonegative integral solutions to $k^2 = 2 n^2 +1$? $n = 2$ and $k = 3$ works, what is the next smallest pair?

Answer 1

First, take the square root of both:

$\sqrt{k^2}=\sqrt{2n^2+1} \iff |k| = \sqrt{2n^2+1}$. In order to get both variables to positive integers, you need to have $2n^2+1$ to be a square number:

If $n = 1 \iff 2n^2+1 = 3$ and is no square number:

$n = 1 \iff 2n^2+1 = 3$

$\color{blue}{n = 2 \iff 2n^2+1 = 9} \rightarrow$ which is a square number

$n = 3 \iff 2n^2+1 = 19$

$n = 4 \iff 2n^2+1 = 33$

$n = 5 \iff 2n^2+1 = 51$

$n = 6 \iff 2n^2+1 = 73$

$n = 7 \iff 2n^2+1 = 99$

$n = 8 \iff 2n^2+1 = 129$

$n = 9 \iff 2n^2+1 = 163$

$n = 10 \iff 2n^2+1 = 201$

$n = 11 \iff 2n^2+1 = 243$

$\color{blue}{n = 12 \iff 2n^2+1 = 289} \rightarrow$ which is also a square number

So, the second value for $n = 12$ and $k=\sqrt{2n^2+1}=\sqrt{289}=17$.