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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!

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    This is not correct as stated. Suppose $a = p$. The residue is a square (it's $0$), but $a$ is not a $p$-adic square. For conditions when there is a square, see Hensel's lemma (Wikipedia, for instance). – 2017-02-28
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    In addition to peter a g's objection we have the special case of $p=2$. A non-zero element $a\in\Bbb{Q}_2$ is a square iff $a=4^m b$ for some integer $m$ and $b\in\Bbb{Z}_2$ such that $b\equiv 1\pmod 8$. Think: squares in $(\Bbb{Z}/8\Bbb{Z})^*$. – 2017-03-01
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    @peterag, one should include a condition that $v_p(a)$ is even, right? – 2017-03-01
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    @Santiago - right... If $\pi$ has order $1$ (e.g., in this case, one can take $\pi = p$ of course), one can write a non-zero $a = \pi^k u$, with $u$ a unit (the expression is unique). Then $a$ is a square iff $k$ is even and $u$ a square in the units. For $p\ne 2$, $u$ is a square iff its residue is a square; Jyrki above explains the case $p=2$. One way to see this (when a unit is a square) is, as advertised above, with Hensel's lemma. One can also do it somewhat directly... – 2017-03-02

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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.

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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.

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    Please see my comment above... – 2017-02-28