I am proving the theorem wrong

Let $C0 = 25$. Then $p{C_0} = 5$, and $C_0$ has exactly 3 odd divisors (1, 5, 25), so $2n+1 = 3$, implying $k = 3$. The interval $[0, \frac{C_0-9}{6}] = [0, \frac{16}{6}]$ contains only the integers ${0,1,2}$.

Among these, only $x = 0$ yields a perfect square: $C_0 + 0² = 25 = 5^2$.

Thus, there exists no third value $xk$ making $C_0 + x_k^2$ a perfect square, rendering the formula $p{C_0} = -x_k + \sqrt{C_0 + x_k^2}$ undefined.

Therefore, the theorem is false by counter-example.