Let $a>0$ be a real cyclotomic number. Is it always possible to solve in cyclotomics the equation $X\overline{X}=a$ ?


Equivalently, one might want to express $a$ as a sum of squares of two real cyclotomics. It is well-known that one square is not always enough.


(If two squares are not enough, then, is there an upper bound?).


**Edit**: $a$ is not only positive, but totally positive (otherwise the answer is No).



is every positive real cyclotomic number the norm of a cyclotomic?
If $a$ is a totally positive real cyclotomic number, then it is a sum of two squares of real cyclotomic numbers.


It suffices to check that the equation $x^2+ y^2 - a z^2=0$ has solutions in real cyclotomic numbers. It has solutions in a particular real cyclotomic number field $F$ if it has solutions everywhere locally. This equation has solutions locally if and only if the quaternion algebra $(a,-1)$ splits locally.


First take the field $F$ generated by $a$. If we adjoin to $F$ a sufficiently large totally real extension of the field of $2$-power roots of unity (sufficiently large depending on the set of ramified places of this quaternion algebra), producing a field $K$, then every ramified place $v$ of this quaternion algebra will not be totally split in $K$; because $K/F$ is a Galois extension of degree a power of $2$, this implies that the quaternion algebra splits over $K\_v$, and so the equation has solutions locally over $K\_v$, thus solutions over $K$.




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How sufficiently large?


For $v$ an odd prime with residue field $q\_v$, it suffices to adjoin the totally real part of the $2^n$th roots of unity for the minimum $n$ such that $q\_v^2 \neq 1 \mod 2^n$. To obtain this, it suffices to have $2^n \geq q\_v^2$.


An odd prime $v$ only ramifies if it divides the numerator or denominator of $a$, in which case $q\_v$ divides the norm of the numerator or denominator, so it suffices to have $2^n$ at least the max of the norms of the numerator and denominator squared.


To handle the even primes $v$, we need to ensure that the image of the inertia group of $2$ acting on the totally real part of field of the $2^n$th roots of unity is strictly larger than the image of the inertia group of $2$ acting on $F$. Since the image of the inertia group acting on the real part of the field of $2^n$th roots of unity is of order the degree $2^{n-2}$, it suffices to take $2^{n-2}> \deg F$.


So it suffices to take $2^n > \max( Na\_1^2, Na\_2^2, 4 \deg F)) $ with $Na\_1, Na\_2$ the norms of the numerator and denominator. We can achieve this with $2^n \leq 2 \max( Na\_1^2, Na\_2^2, 4 \deg F)$, which is an extension of $F$ of degree $\leq ( Na\_1^2/2, Na\_2^2/2, 2\deg F)$.


