$p$-adic perfect square condition. 
Let $a$ be an integer $p$-adic number for $p\ge2$. Prove that $a$ is a perfect square iff $\exists$ such integer $x$ that $a \equiv x^2 mod$ $p$.

It is easy to prove $\Rightarrow$: if $a$ is a perfect square then $\exists$ such $b = (...,b_2, b_1, b_0)$ that $b^2 = a$, then $x = b_0$.
But I have no idea about $\Leftarrow$. Thanks!
 A: Easiest to specify the squares among the units. Remember that, using standard $p$-ary notation, a $p$-adic unit is of form $\dots a_4a_3a_2a_1a_0;$ with $a_0\ne0$.
The situation is special for $p=2$, since it’s not enough for $a_0$ to be a square. Rather, the $2$-adic units that are squares are those that are $\equiv1\pmod8$, in other words the right-hand end must be $\dots a_4a_3001;$ (note that $3=11;$, $5=101;$, and $7=111;$ all are non-squares in $\Bbb Q_2$, but $-7=\dots1111001;$ is a square).
For primes bigger than $2$, what you say is (almost) right: for a unit, the $a_0$ must be a nonzero square modulo $p$.
For nonzero nonunits $z$, you may always write $z=p^ru$, with $r\in\Bbb Z$ and $u$ a $p$-adic unit. Then $z$ is a square if and only if $r$ is even and $u$ is a square unit.
A: Consider $y=y_0+y_1p+y_2p^2+\cdots $ then $y^2\equiv y_0^2\pmod{p} $.
Now its obvious that if there exist integer $x$ such that $a\equiv x\pmod{p}$ then $a $ is a perfect square. In fact $a\equiv x\pmod{p}$ leads to the two solutions of the equation.
