If $x^2 \equiv a \pmod{p}$, can we use Jacobi symbol to determine the equation has solution? To determine a quadratic congruence equation has any solution, we have to evaluate 
$$\bigg( \dfrac{a}{p} \bigg)$$, So can we apply the algorithm of Jacobi symbol to evaluate this?
If yes, what's the easiest way to evaluate this symbol? Let's not take efficiency into account because I don't have a computer in my exam, so I just want to learn the simplest way which I will be able recall it easily during the exam. 
For example, evaluate $$\bigg( \dfrac{2819}{4177} \bigg)$$
$$(\dfrac{2819}{4177}) = (\dfrac{4177}{2819}) \text{ because } \frac{(2819-1)(4177-1)}{4} = \text{ even } $$ 
$$= (\dfrac{1358}{2819}) = (\dfrac{2}{2819}) \cdot (\dfrac{679}{2819})$$
$$= (-1) \cdot (\dfrac{679}{2819}) \text{ because } \frac{2819^2 - 1}{8} = \text{ odd }$$
$$= (-1) \cdot (-1) \cdot (\dfrac{2819}{679}) = (\dfrac{2819}{679}) \text{ because } \frac{(m-1)(n-1)}{4} = \text{ odd }$$
$$= (\dfrac{2819}{679}) = (\dfrac{103}{679}) = (-1) \cdot (\dfrac{679}{103}) \text{ because } \frac{(103-1)(679-1)}{4} = \text{ odd }$$
$$= -(\dfrac{61}{103}) = -(\dfrac{103}{61}) \text{ because } \frac{(61-1)(103-1)}{4} = \text{ even }$$
$$= -(\dfrac{42}{61}) = -(\dfrac{2}{61}) \cdot (\dfrac{21}{61}) = -(\dfrac{21}{61}) \text{ because } \frac{61^2 - 1}{8} = \text{ even }$$
$$= -(\dfrac{61}{21}) \text{ because } \frac{(21-1)(61-1)}{4} = \text{ even }$$
$$= -(\dfrac{19}{21}) = -(\dfrac{21}{19}) \text{ because } \frac{(19-1)(21-1)}{4} = \text{ even }$$
$$= -(\dfrac{2}{19}) = (-1) \cdot (-1) = 1 \text{ because } \frac{19^2-1}{8} = \text{ odd }$$.
Since in the book example, the author often skipped many steps, so I just want to make sure that I understand it correctly. 
Any suggestion and idea would be greatly appreciated.
Thank you,

It looks to me like what you have done is fine. Did you check to see whether 4177 is prime? If it isn't, then Jacobi symbol 1 doesn't guarantee quadratic congruence solvable, e.g., $x^2\equiv2\pmod{15}$ has no solution even though the symbol is 1. 
