Find prime number which satisfies $p \in {\Bbb{{Q}_2}^{\times}}^2-{{\Bbb{Q}_2}^{\times}}^4$

Question

What is an minimal prime number $p$ which satisfies $p \in {\Bbb{{Q}_2}^{\times}}^2-{{\Bbb{Q}_2}^{\times}}^4$?

For $a \in \Bbb{Z}$, if $a≡1,3,4,7$mod8, then for each $a$, $\Bbb{Q}_2$ is $\Bbb{Q}_2(\sqrt{a})$ is isomorphic to $\Bbb{Q}_2,\Bbb{Q}_2(\sqrt{3}),\Bbb{Q}_2(\sqrt{5}),\Bbb{Q}_2(\sqrt{7})$.

So I searched $p=17,41,73,\ldots$, which are primes which is $1\bmod8$.

For $p=17$, ${\rm ord}_2(f(3))=6>2{\rm ord}_2(f'(3))＝4$, thus by Hensel lemma, $17^{1/4}\in \Bbb{Q}_2$. So $p=17$ is not ok.

In such a way, can I find titled $p$ ? Or should I choose another way or computer ? Computer method answer is also appreciated. Thank you for your help.

Answer 1

We have the multiplicative decomposition $\mathbf Z_2^\times = \{\pm 1\} \times (1 + 4\mathbf Z_2)$, so $(\mathbf Z_2^\times)^2 = (1 + 4\mathbf Z_2)^2$. Using the $2$-adic exponential, $$ 1 + 4\mathbf Z_2 = \exp(4\mathbf Z_2) \Longrightarrow (1+4\mathbf Z_2)^2 = \exp(8\mathbf Z_2). $$ The $2$-adic exponential and logarithm are isometries between $4\mathbf Z_2$ and $1 + 4\mathbf Z_2$, so $\exp(8\mathbf Z_2) = 1 + 8\mathbf Z_2$. Similarly, $$ (1+4\mathbf Z_2)^4 = \exp(16\mathbf Z_2) = 1 + 16\mathbf Z_2. $$ So the $u \in \mathbf Z_2^\times$ that are squares and not fourth powers are exactly the $u$ where $u \equiv 1 \bmod 8$ and $u \not\equiv 1 \bmod 16$, which is equivalent to $u \equiv 9 \bmod 16$. The first few such $u$ that are prime numbers are $41$, $73$, $89$, $137$, and $233$.

Answer 2

The squares in $\mathbb{Q}_2^{\times}$ are exactly the elements of the form $1+8u$ with $u \in \mathbb{Q}_2$. Hence such an element is a fourth power iff the square root of $1+8u$ is itself of the form $1+8u'$.

Let's see if we can do this. The square root of $1+8u$ can be obtained by evaluating the binomial series $$ (1+8u)^{1/2}=\sum_{n=0}^\infty \binom{1/2}{n} (8u)^n = 1+4u-8u^2+\ldots $$ Now the $2$-adic valuation of the binomial coefficient $\binom{1/2}{n}$ equals $-n-\operatorname{ord}_2(n!)$, which by Legendre's formula equals $$ -n-\left\lfloor \frac{n}{2} \right\rfloor-\left\lfloor \frac{n}{4} \right\rfloor-\left\lfloor \frac{n}{8} \right\rfloor-\ldots \geq -n \left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right)=-2n, $$ which means that the coefficient of the $u^n$ term in the binomial formula has valuation $\geq n$ (this also shows that the series is convergent by the way). In particular, none of the terms of degree $3$ or higher will contribute anything to the sum modulo $8$.

We may therefore conclude that the square root of $1+8u$ is congruent to $1+4u$ modulo $8$, and this number is therefore itself a square if and only if $u=2u'$ for some $u'$. Hence the fourth powers are those elements which are of the form $1+16u'$, and the squares which are not fourth powers are those elements which are of the form $9+16u'$. In particular, the (rational) primes $41$ and $73$ would be examples of such elements.