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Do the p-adic integers, rationals, or any of its algebraic closures have interesting, unique properties for specific p or subsets of the prime numbers(i.e. the Fibonacci primes or twin primes)? Are there any open problems that have to do with such properties? The p-adic numbers are interesting for many reasons, but this would certainly take it to another level.

Adam French
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    $\mathbb{Q}_2$ is quite different from $\mathbb{Q}_p$ for odd $p$. For example, radius of convergence of exponential function has a slightly different form, and the unit group of $\mathbb{Z}_p$ are also slightly different. These thing happen because $p=2$ is small, not because $p=2$ is even. – Seewoo Lee Sep 22 '20 at 00:27

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Each $p$-adic field $\Bbb Q_p$ has its own peculiar properties. For instance, when $p\equiv1\pmod4$, $\Bbb Q_p$ contains a square root of $-1$. More generally, the group of roots of unity in $\Bbb Q_p$ is of order $p-1$, except when $p=2$. In $\Bbb Q_2$, the roots of unity are just $\{\pm1\}$.

Lubin
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If you want to use $p$-adic numbers to determine all the integral solutions to some equation (e.g., by Skolem's method or Strassmann's theorem) then you bound the number of integral solutions by bounding the number of $p$-adic integral solutions and that bound varies with $p$.

For some $p$ you may get an upper bound on the number of $p$-adic integral solutions that exceeds the number of expected integral solutions; perhaps the equation has solutions in $\mathbf Z_p$ that are not in $\mathbf Z$. By changing $p$ to another prime you may be fortunate to get a small upper bound on $p$-adic integral solutions for the new prime that matches the known number of integral solutions and then you're done.

For example, consider the problem of showing the sequence $\{a_m\}$ defined by the linear recursion $$ a_m = 2a_{m-1} - 3a_{m-2} $$ and initial conditions $a_0 = a_1 = 1$ has $a_m = 1$ only for $m = 0, 1$, and 5. I first came across this question as the math.stackexchange post Finding non-negative integers $m$ such that $(1 + \sqrt{-2})^m$ has real part $\pm 1$., where I solved it by using Strassmann's theorem in $\mathbf Q_2(\sqrt{-2})$. Later I realized the problem can be solved by working in $\mathbf Q_{11}$, which has square roots of $-2$ and thus avoids the need to work in extension fields of some $\mathbf Q_p$: see Theorem 1.1 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/strassmannapplication.pdf. The actual solution there in $\mathbf Q_{11}$ is in Section 5. In Section 6 I use $\mathbf Q_{11}$ to show $a_m = -5$ only for $m = 3$ and $a_m = 23$ only for $m = 6$, but to show $a_m = -241$ only at $m = 10$ there is a problem with using $\mathbf Q_{11}$: two additional solutions appear in $\mathbf Z_{11}$ that should not correspond to an $m \in \mathbf Z^+$ for which $a_m = -241$.
So let's change the setting to $\mathbf Q_p$ for another prime $p$ besides $11$. The primes $p > 3$ for which $\mathbf Q_p$ contains $\sqrt{-2}$ start off as $11, 17, 19, 41, \ldots$. In $\mathbf Q_{17}$ and $\mathbf Q_{19}$ the upper bound on the number of $p$-adic integral solutions is $2$ rather than $1$, so it's still too high. In $\mathbf Q_{41}$ the upper bound on $41$-adic integral solutions is $1$, so finally we are done since that matches the number of known integral solutions.

KCd
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