<s>[INST] How can I solve $y^4 = 5 \pmod{11\times19}$ with legendre? Solve $y^4 = 5 \pmod{11\times19}$
I'm trying to let $y^2=A$ then $A^2=5 \pmod{11\times19}$. 
And solve this problem then $A= 104,-104,28,-28 \pmod{11\times19}$
Then should I solve this problem for all this four $A$ ?
I want to solve this easily with Legendre.
(I think I can use this thing that $(\frac{100}{14})=1$ and $(\frac{104}{19})=1$ , $(\frac{28}{11})=-1$, $(\frac{14}{11})= 1$.)
All fractions are translated as Legendre symbol.
 [/INST] Separating out the $\bmod 11$  and $\bmod 19$ strands of the solution (just looking up to $\lfloor n/2 \rfloor$, we can use symmetry after that):
\begin{array}{c|c|c} n & n^2 \bmod 11 & n^4 \bmod 11 & n^2 \bmod 19 & n^4 \bmod 19\\ \hline
1 & 1 & 1 & 1 & 1 \\
2 & 4 & \color{red}
        5 & 4 & 16 \\
3 & 9 & 4 & 9 & \color{red}
                 5 \\
4 & 5 & 3 & 16 & 9 \\
5 & 3 & 9 & 6 & 17 \\
6 &   &   & 17 & 4 \\
7 &   &   & 11 & 7 \\
8 &   &   & 7 & 11 \\
9 &   &   & 5 & 6 \\
\end{array}
This gives us $y \equiv \pm 2\equiv (2,9) \bmod 11$ and  $y \equiv \pm 3\equiv (3, 16) \bmod 19$ 
\begin{align}
y=19k+3 &\equiv \pm2 \bmod 11\\
8k &\equiv -1 \bmod 11 \tag{from $+2$}\\
k = 4 \implies y &=76+3 = 79 \text{ is a solution}\\
8k &\equiv -5  \bmod 11 \tag{from $-2$}\\
k = -4\times -5 \equiv 9 \implies y &= 171+3 = 174 \text{ is a solution}\\
\end{align}
So $ y \equiv (\pm 79, \pm 174) \equiv (35, 79, 130, 174) \bmod 209$ are the solutions 

The numbers are small enough that we could also just step through the possibilities:
$\{2,9,13,20,24,31,\color{red}{35},42,46,53,57,64,68,75,\color{red}{79},86,90,97,101,\ldots\}$
$\{3,16,22,\color{red}{35},41,54,60,73,\color{red}{79},92,98,\ldots\}$
Then $ y \equiv (\pm 35, \pm 79) \equiv (35, 79, 130, 174) \bmod 209$
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