Trick(y) Question

I thought that $5$ was a Heegner number, but apparently not.

$\mathbb{Z}[\sqrt{-5}]$ does not have unique factorisation; this can be easily seen by the two different factorisations of $6$ :

$6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$

Now, a prime $p$ is a non-zero non-unit element in an integral domain such that if $p$ divides $ab$ , then $p$ divides $a$ or $b$ . However, clearly no such integer divides $2$ , $3$ , $1+\sqrt{-5}$ or its conjugate. As such, $6$ as no prime factors.

If the domain $\mathbb{Z}[\sqrt{-d}]$ has unique factorisation such as in the case $d = 7$ or $11$ or most famously $163$ , then all primes in $\mathbb{Z}$ are primes in $\mathbb{Z}[\sqrt{-d}]$ (your usual).

Correction : See Otto Bretscher's comment.