Prove that if neither $a$ nor $-a$ are squares modulo $d$, then no integral ideal in $K$ of norm $a$ is principal. Let $K=\mathbb{Q}(\sqrt{d})$ where $d \neq 0,1$ is a squarefree integer.
Let $a \in \mathbb{Z}$. Prove that if neither $a$ nor $-a$ are squares modulo $d$, then no integral ideal in $K$ of norm $a$ is principal.
Hint: Consider cases disk $K=d$ and disc $K=4d$.
For the case disk $K=d$, we have $\mathbb{Z}_k=\mathbb{Z}[\alpha]$ where $\alpha=\frac{1+\sqrt{d}}{2}$. The norm of a general element $x+y\alpha$ is $(x+y/2)^2-d(y/2)^2$.
I suppose $(d/a)=-1$. I am now stuck. Any help?
 A: It doesn't actually matter if $a$ is prime in $\mathbb Z$ or not. Under the conditions you've specified, $x^2 - dy^2 = \pm a$ has no solutions, which means that in the domain at hand, $N(z) = \pm a$ is impossible. Clearly $N(\pm a) = a^2$, and therefore $N(\langle a \rangle) = a^2$ as well.
We don't really need to worry about the cases $4a$, $-4a$ either. That's because $4$ is pretty much always a square in whatever ring it exists in. Thus, if $a$ is a square, then $4a$ is as well.
So, for example, in $\mathcal O_{\mathbb Q(\sqrt{65})}$, since neither $3$ nor $62$ are squares modulo $65$, we don't actually have to check that $12$ and $53$ aren't either.
A: So $a$ is a purely real, rational integer. Is $a$ a prime in $\textbf Z$? Let's say that it is. It's certainly irreducible in $K$, since, as you've figured out, $$\left(\frac{d}{a}\right) = -1.$$ Given that the relation $x^2 \equiv d \pmod a$ has no solutions, that guarantees that $a$ is also prime in $K$. This means that a number with a norm of $a$ is nonexistent in $K$, since $N(a) = a^2$.
Concrete example: $d = 10$, $a = 7$. Try to solve $x^2 \equiv 7 \pmod{10}$, you can't. Nor can you solve $x^2 \equiv 3 \pmod{10}$. More importantly, though, $x^2 \equiv 10 \pmod 7$ has no solution either. Therefore no number $x \in \textbf Z[\sqrt{10}]$ satisfies $N(x) = \pm 7$, and $N(7) = 49$.
Compare $a = 3$. You also come up empty on $x^2 \equiv 3 \pmod{10}$ and $x^2 \equiv 7 \pmod{10}$. But $x^2 = 10 \pmod 3$ does have solutions, the most obvious one being $x = 1$, leading you to find the ideals $\langle 3, 1 - \sqrt{10} \rangle$ and $\langle 3, 1 + \sqrt{10} \rangle$, neither of which is principal.
A: Your hypotheses are a bit muddled. On the one hand you assume that $\pm a$ is not a square mod $d$ , and on the other you write $(d/a) = -1$, i.e., if $a$ is prime, that $d$ is not a square mod $a$. So let us try to clear up the necessary requirements as we go along. 
First you speak of ideals $I$ of the ring of integers $A$ of your quadratic field $K = \mathbf Q (\sqrt d)$ with norm $a \in \mathbf Z$. There are at least two definitions of $N(I)$. The first one is $N(I)$ is the order of the quotient $A/I$, which implies that $a$ must be a natural integer.
If you consider $a \in \mathbf Z$, then the definition should be: $N(I)$ and $a$ generate the same ideal in $\mathbf Z$, in which case the sign of $a$ becomes irrelevant. But then you could also consider the ideals generated in $A$, in which case the definition becomes $aA$ is the product of all the distinct Galois conjugates of $I$, see the answer to this question about norms and ideals.
Let us stick to the latter definition. Because the norm is multiplicative, we can assume that $a$ is a rational prime $p$. Then the theory of the decomposition of primes in a quadratic field tells you that $p$ is a norm iff $p$ splits, i.e. $pA = P.P'$, where $P$ and $P'$ are two conjugate prime ideals of $A$, if and only if $(d/p) = 1$ if $p$ is odd, and $d \equiv 1 \bmod 8$ if $p=2$. At this stage :


*

*if your main hypothesis is $(d/p) = -1$ for $p$ odd, then the case $p = 2$ requires further examination. The point is that $A$ does not necessarily coincide with $B = \mathbf Z [\sqrt d]$ (see the hint on $\textrm{disc}(K)$). An easy calculation shows that $A/pA = B/pB$ if $p$ is odd, but the case $p = 2$ requires more tedious computation to obtain the criterion that $2$ splits if and only if $d \equiv 1$ mod $8$. Anyway, it remains in this case to check that the prime ideals above $2$ cannot be principal. 

*if your main hypothesis is that $\pm p$ is not a square mod $d$, then a direct calculation of norms of elements solves the problem, as in the shortcut given by @Lisa. But of course the long detour that we made was only meant to clear up your assumptions and understand what is going on. 
