Suppose $p$ is a prime number such that $p\equiv 7 \pmod{12}$.


Since $p \not \equiv 1 \mod{4}$, the ring of integers of $\mathbb{Q}(\sqrt{p})$ is $\mathbb{Z}\oplus\mathbb{Z}(\sqrt{p})$ with fundamental unit of the form $a+b\sqrt{p}$, where $a, b > 0$


It is of the norm $+1$, because $a^2 - pb^2 \equiv -1 \pmod{4}$ doesn't have solutions when $p\equiv -1 \pmod{4}$.


It is easy to see that $a\equiv \pm 1 \pmod{3}$, since if $3|a$, then $a^2 - pb^2 \equiv - b^2 \not\equiv 1 \pmod{3}$



> 
> **A computer-checked observation**. For all said $p$ such that $p<20000$ the following holds: $a\equiv -1 \pmod{3}$
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> 
> 


That, I think, is rather surprising. I tried proving that $a\equiv -1$ for the case when the period of continued fraction for $\sqrt{p}$ is of length $4$, but got nowhere.


The claim is false when $p$ is not prime, for example $\mathbb{Q}(115)$ has the fundamental unit $1126+105\sqrt{115}$. It is also false when we drop modulo conditions, say, in $\mathbb{Q}(\sqrt{11})$ the fundamental unit is $10+3\sqrt{11}$


What is known about coefficients of the continued fraction for $\sqrt{p}$ ? I only know that it is of the form $[a\_o; \overline{a\_1, a\_2, ..., a\_2, a\_1, 2a\_0}]$ and that when the class number of the field is 1, the sum $a\_1 - a\_2 + a\_3 - ... \pm 2a\_0$ is divisible by 3 (the latter is a corollary from a theorem by Don Zagier).


All that cannot even preclude the case when all coefficients of the continued fraction are divisible by 3.


Any hints are welcome.



First coefficient of totally positive fundamental unit modulo 3
By Theorem 1.1 of Zhang-Yue: Fundamental units of real quadratic fields of odd class number, J. Number Theory 137 (2014), 122-129, we have that $a\equiv 0\pmod{2}$. From here it is a simple matter to prove that $a\equiv 2\pmod{3}$, hence in fact $a\equiv 2\pmod{6}$. To see this, let us write
the unit equation as
$$(a-1)(a+1)=pb^2.$$
The left hand side is divisible by $3$, hence so is $b$. The factors on the left hand side are odd and coprime, so they are of the form $pb\_1^2$ and $b\_2^2$ in some order, where $b=b\_1b\_2$. Note that exactly one of the $b\_i$'s is divisible by $3$. Therefore, adding the factors of the left hand side, we get modulo $3$ that
$$2a=pb\_1^2+b\_2^2\equiv b\_1^2+b\_2^2\equiv 1\pmod{3}.$$
Hence $a\equiv 2\pmod{3}$ as claimed.


