The equation $x^2+P=0$, where $P$ is an integer, has 2 integer solutions for $x$. Which of the following is a possible value of $P$ ?

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We can inspect each option:

$$x^2-48 =0$$ $$x^2-36=0$$ $$x^2-10 = 0$$ $$x^2+36=0$$ $$x^2+48=0$$

Only the second one, when $x=6$ and $x=-6$, seems appropriate.

$$x^2+P = 0$$ $$x^2= -P$$ $$x = \pm \sqrt{-P}$$

We will obtain two solutions if P is a negative perfect square.