Parametrization of Pythagorean triplets

Consider the number to be filled in the parentheses be $x^2$ and obtained number be $p^2$ . Then the equation can be written as $\begin{aligned}&169+x^2 = p^2 \\& p^2-x^2 = 169\end{aligned}$ Noting that $169$ is perfect prime square number of $13$ so it will be coprime integers expect the numbers in $13^k, k\in\mathbb N$ , $1$ and numbers in it multiple integral and $13$ , $1$ are only the factor of $169$ less than it, then we have following possible cases for above equation. $\begin{cases} p^2-x^2 = 13^2 \Rightarrow (p-x)(p+x) = 13\times 13 \\p^2-x^2 =169\Rightarrow (p-x)(p+x) = 1\times 169 \end{cases}$ Solving the equations above we get $x = 0\space\text{or}\space 84$ . Since the positive is $84$ So , $84$ is the required positive integer.

• The difference of two consecutive perfect squares is always odd. $(n+1)^2$ - $n^2$ =2n+1. In other words, $n^{2}$ + (2n+1) = $(n+1)^2$ . for example, $10^2$ + $(2*10+1)$ = $11^2$ . Therefore, in this question, $3^2$ + $4^2$ + $12^2$ = 169 which is of the form ( $2*84 +1$ ) + $84^2$ = $85^2$ . Hence, the answer is 84