<s>[INST] 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,
 [/INST] 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. 
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