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happy new year
I have this statement: "By quadratic reciprocity there are the integers $a$ and $b$ such that $(a,b)=1$, $(a-1,b)=2$, and all prime $p$ with $p\equiv a$ (mod $b$) splits in $K$ (where $K$ is a real quadratic field)".
I have tried with many properties of quadratic reciprocity but couldn't even get to the first conclusion.
Thank you very much in advance, for any idea or advice for approach the problem
The point is that (because of quadratic reciprocity), if $D$ is the discriminant of $K$, then splitting in $K$ is determined by the congruence class of $p$ mod $D$ (more precisely, by the Jacobi symbol mod $D$). If $D$ is even, take $b = D$, if $D$ is odd, take $b = 2D$.
The problem is now to find a residue class in $(\mathbb Z/b)^{\times}$ such that the Jacobi symbol of $a$ mod $D$ is trivial (this is easy; half of the elements in $(\mathbb Z/b)^{\times}$ has this property), and such that $(a-1,b) = 2$.
For simplicitly, suppose that $D = q$, an odd primes. Then you can take $a$ to be any residue class with trivial Legendre symbol such that $a-1$ is not zero mod $q$. Since $q \equiv 1 \bmod 4$ (it is a fundamental discriminant), it is at least $5$, so there is a quadratic residue mod $q$ beside $1$. Let $a$ be any odd representative of this quadratic residue. Then $(a,2q) = 1,$ $(a-1,2q) = 2$, and if $p \equiv a \bmod b$, then $p$ splits in $\mathbb Q(\sqrt{q}).$
Presumably the general case can be handled in a similar way, using the Chinese Remainder theorem (applied to the prime factorization of $D$).
Edited to address some bizarrely horrible errors in the first version.
Here's a simple case from which it should not be too hard to generalize. Suppose that $K$ is a quadratic field of prime discriminant $q$. Since $q\equiv 1\pmod{4}$, note that a prime $p$ splits in $K$ if and only if $\left(\frac{p}{q}\right)=1$.
Let $b=2q$ and choose an integer $a\not\equiv 1\pmod{q}$ which is an odd quadratic residue mod $q$. Such a thing exists since there are $\frac{q-1}{2}\geq 2$ (since $q\geq 5$) residues mod $q$, and if $a'$ is any not-conguent-to-1 residue mod $q$, then one of $a'=a$ and $a'=q+a$ gives you an odd residue. For example, when $q=5$, take $a'=4$ and then $a=9$. (Actually, $p=5$ is the only example where you have to add $q$: for all other $q$ there is an odd quadratic residue $a$ in the range $2\leq a\leq p-1$).
Now $a$ is odd and not divisible by $q$, so $(a,b)=(a,2q)=1$, and since $a\not\equiv 1\pmod{q}$, we also have $(a-1,b)=2$. Finally, if $p\equiv a\pmod{b}$, then $p\equiv a\pmod{q}$, so $\left(\frac{p}{q}\right)=\left(\frac{1}{q}\right)=1$, and $p$ splits in $K$.
Just some minor commentary on where this came from: Your $\gcd$ conditions force $b$ to be even, and for a congruence-mod-b condition to determine splitting in $K$, your $b$ needs to be a multiple of the discriminant (in this case, $q$), and preferrably as small a multiple as possible to prevent extra congruence classes from slipping in. The value of $b=2q$ satisfies all of these requirements, and since clearly one must choose $a$ to be an odd quadratic residue mod $q$, you're left with essentially the above construction.
