Consecutive Adding

We can rewrite the equation as $2A = (2n - 1)x + x^{2} \Longrightarrow 2^{b+1} = x(2n + x - 1)$ .

Now as $x(2n + x - 1)$ is an integer we must have that $b + 1 \ge 0$ , and thus that $2^{b+1} \ge 1$ . But as $x \gt 1$ , by the Fundamental Theorem of Arithmetic we must have that $x = 2^{a}$ for some positive integer $a$ , ensuring that $x$ is even, and thus also that $2n + x - 1$ is a non-negative integer power of $2$ . But since $2n + x - 1$ is necessarily odd, the only way for it to be such a power of $2$ is for $2n + x - 1 = 1$ . However, with $n,x$ being positive we have that $2n + x - 1 \ge 2$ , and thus there are no ordered pairs $(n,x)$ that satisfy the given equation.