Heegner Number Modular Arithmetic Why are all Heegner numbers > 8 (up to 163) all equal 3, mod8
i.e Let $H$ denote a Heegner number. When $H$ > 8
$$H \equiv 3\mod8$$
 A: You basically show that every other residue $\bmod 8$ leads to nonunique factorizations.
First, composite numbers such as $6$ and $15$ are knocked out because the composite number itself will have a nonunique factorization. For instance, $15$ in the integers of $\mathbb Q[\sqrt{-15}]$ is both $3×5$ and $(-1)(\sqrt{-15})^2$, and these factors cannot be reduced because there are no elements with norm $3$ or $5$.
That leaves only $1$ and primes, thus above $2$ only odd numbers. Next, propose a Heegner number of the form $4k+1$, like $5$ or $13$. Then in $\mathbb Z[\sqrt{-(4k+1)}]$ we have
$4k+2=2(2k+1)=(1+\sqrt{-(4k+1)})(1-\sqrt{-(4k+1)})$
where it is understood that $2k+1$ can be factored further. But for this factorization to be unique, given that $(1\pm\sqrt{-(4k+1)})/2$ is not an algebraic integer, $2$ must he reducible and it isn't, because there are no elements in the integer domain with norm $2$.
So Heegner numbers above $2$ are all one less than a multiple of $4$, meaning they have residue $3$ or $7\bmod 8$. For residue $7$, try to factor $2k+2$ where the proposed Heegner number is $8k+7$. Even though you have algebraic integers of the form
$(m+\frac12)+(n+\frac12)\sqrt{-(8k+7)}$
available, you again find that the factor $2$ is irreducible, forcing the factorization of
$2k+2=2(k+1)=(\frac12+\frac12\sqrt{-(8k+7)})(\frac12-\frac12\sqrt{-(8k+7)})$
to be nonunique, except for the one case $k=0$ (where $(1\pm\sqrt{-7})/2$ are algebraic integers with norm $2$). Thus $7$ itself is possibly a Heegner number on this line of reasoning, but no larger number congruent to it $\bmod 8$.
So by process of elimination, the only possible Heegner numbers must be $1,2,7$ and numbers $\equiv3\bmod 8$.
