<s>[INST] How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution? 
Any help would be appreciated!
 [/INST] The numbers are small, so general techniques are not necessary. However, we describe, in this particular case, the method of Hensel Lifting. 
It is clear that the solutions modulo $7$ are $x\equiv \pm 3\pmod{7}$. We lift the solution $x\equiv 3\pmod{7}$ to a solution of $x^2\equiv 2\pmod{7^2}$. 
Any solution to $x^2\equiv 2\pmod{7^2}$ that is congruent to $3$ modulo $7$ has shape $x=3+7t$. Square. Modulo $7^2$, the result is congruent to $9+(2)(3)(7)t\pmod{7^2}$. We want this to be congruent to $2$ modulo $7^2$, so we want
$$9+(2)(3)(7)t\equiv 2\pmod{7^2}.$$
A little manipulation turns this to
$$(2)(3)t\equiv -1\pmod{7}.$$
Thus $t=1$ works, and we have
$$x\equiv 10\pmod{7^2}.$$
Now lift this solution to a solution of $x^2\equiv 2\pmod{7^3}$. We look for a solution of the shape $x\equiv 10+7^2t\pmod{7^3}$. 
Squaring, we get
$$100+(2)(10)(7^2)t\equiv 2\pmod{7^3},$$
which simplifies to 
$$(2)(10)t\equiv -2\pmod{7}.$$
A solution is $t\equiv 2\pmod{7}$. That gives solution $x\equiv 10+2(49)\pmod{7^3}$.
By general theory, the only solutions are therefore $x\equiv \pm 108\pmod{7^2}$.
Remark: If we wished to, we could continue, and lift to a solution modulo $7^4$, $7^5$, and so on. 
Hensel lifting is an important general technique for solving polynomial congruences modulo prime powers.  
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