Let $n$ be the number of points $P$ interior to the region bounded by a circle with radius $1$, such that the sum of squares of the distances from $P$ to the endpoints of a given diameter is $3$. Then $n$ is:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 4\quad \text{(E) } \infty$
Let $A$ and $B$ be points on diameter.  Extend $AP$, and mark intersection with circle as point $C$.
Because $AB$ is a diameter, $\angle ACB = 90^\circ$.  Also, by Exterior Angle Theorem, $\angle ACB + \angle CBP = \angle APB$, so $\angle APB > \angle ACB$, making $\angle APB$ an obtuse angle.
By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines), $AP^2 + BP^2 - 2 \cdot AP \cdot BP \cdot \cos{\angle APB} = 4$.  Since $AP^2 + BP^2 = 3$, substitute and simplify to get $\cos{\angle APB} = -\frac{1}{2 \cdot AP \cdot BP}$.  This equation has infinite solutions because for every $AP$ and $BP$, where $AP + BP \ge 2$ and $AP$ and $BP$ are both less than $2$, there can be an obtuse angle that satisfies the equation, so the answer is $\boxed{\textbf{(E)}}$.