Let $p$ be a prime number, such that $p\equiv 1\pmod 8.$ and $K=\mathbb Q(\sqrt p).$ Can Someone help me to prove this : $$\epsilon=a+b\sqrt p\hspace{4mm}a,b\in \mathbb Z\hspace{2mm}\text{is a fundamental unit of }\hspace{4mm} K\Rightarrow a\equiv 0\pmod 4\hspace{2mm} \text{and}\hspace{2mm} N_{K/\mathbb Q}(\epsilon)=-1.$$

Thank you.

• I think that you must us Pell's equation. – Med Jan 14 '14 at 16:38
• What about $p = 17$ and $\epsilon = 4+ \sqrt{17}$? The norm fits, but $a \not \equiv 1 \bmod 4$. – benh Jan 14 '14 at 19:01
• @user44319 They are inverses, so both generate the units up to a sign. I thought the convention was to choose the generator $>1$. Anyways, neither the one nor the other suffice the congruence $a \equiv 1 \bmod 4$. – benh Jan 14 '14 at 20:40
• @benh , is a typo, thanks. – Edgar Jan 14 '14 at 22:44
• @Edgar what are $a$ and $b$ here? – Konstantin Ardakov Jan 14 '14 at 23:25

See Exercise 16 of these notes for hints on how to find some $u,v \in \mathbb{Z}$ with $u^2 - pv^2 = -1$ (the condition $p \equiv 1 \mod 4$ is used there).

Now $u + v \sqrt{p} = \pm \epsilon^m$ for some $m \in \mathbb{Z}$. If $N_{K/\mathbb{Q}}(\epsilon) = 1$ then $-1 = N_{K/\mathbb{Q}}(u + v \sqrt{p}) = N_{K/\mathbb{Q}}(\pm \epsilon^m) = 1$, a contradiction.

So $N_{K/\mathbb{Q}}(\epsilon) = a^2 - pb^2 = -1$.

Since $p \equiv 1 \mod 8$, we can reduce this equation $\mod 8$ to get $a^2 \equiv b^2 - 1 \mod 8$.

The values of the left hand side can only be $0,1,4 \mod 8$. The values of the right hand side can only be $7,0,3 \mod 8$. The only overlap is $0$, so $a^2 \equiv 0 \mod 8$.

Therefore $a \equiv 0 \mod 4$.

• Konstantin thank you so much. – Edgar Jan 15 '14 at 3:03