Quadratic residues and composite moduli

The point of the problem is to avoid thinking that the answer is (3) and (4).

It's not hard to check that $73$ is not a quadratic residue mod $5,$ and an easy computation with Legendre symbols (and quadratic reciprocity ) shows that it is not a quadratic residue mod $83$ either: $\begin{aligned} \left( \frac{73}{83} \right) &= \left( \frac{83}{73} \right) \\ &= \left( \frac{10}{73} \right) \\ &= \left( \frac{2}{73} \right) \left( \frac{5}{73} \right) \\ &= \left( \frac{5}{73} \right) \\ &= \left( \frac{73}{5} \right) \\ &= \left( \frac35 \right) = -1. \end{aligned}$ (Here we've used that $2$ is a square mod an odd prime $p$ if and only if $p \equiv \pm 1 \pmod 8,$ which is the so-called second supplement to quadratic reciprocity.)

So the Jacobi symbol $\left( \dfrac{73}{415} \right)$ equals the product of $\left( \dfrac{73}{5} \right)$ and $\left( \dfrac{73}{83} \right),$ which is $-1 \cdot -1 = 1.$ Hence statements (1) and (2) are false and statement (4) is true.

But, unlike the Legendre symbol, the Jacobi symbol being equal to $1$ does not imply that $73$ is a square mod $415.$ Indeed, if $x^2 \equiv 73 \pmod{415},$ then $x^2 \equiv 73 \pmod 5,$ which is impossible. In fact, the Chinese Remainder Theorem shows that it's a square mod $415$ if and only if it's a square mod $5$ and mod $83.$ So it's not enough that the product of the Legendre symbols is $1$ --all the Legendre symbols have to be $1$ individually!

This was the first question I was asked as a graduate student during my oral qualifying examination ("is 73 a square mod 415?"); I believe it was asked as a very gentle way to warm me up and calm me down, since it was much easier than the questions that came afterward!