<s>[INST] Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$? I dont know if this is correct...
$(\frac{21}{-17}) = (\frac{3}{-17}) * (\frac{7}{-17})$
for $(\frac{3}{-17}) = (-1) (\frac{-17}{3})(\text{Quadratic reciprocity})= (\frac{17}{3}) = (\frac{2}{3}) = -1$
for $(\frac{7}{-17}) = (-1) (\frac{-17}{7})(\text{Quadratic reciprocity})= (\frac{17}{7}) = (\frac{3}{7}) = (-1)(\frac{7}{3})(\text{Quadratic reciprocity}) = (\frac{2}{3}) = -1$
insert, we get:
$(-1) * (-1) =1$ and we have integer solutions?
 [/INST] $$x^2-21=17y$$
Let $x=3k+2$, $k\in\mathbb Z^{+}∪{0}$, we have
$$\begin{align}x^2-21&=(3k+2)^2-21\\
&=9k^2+12k-17\end{align}$$
Then let, $k=17m, m\in\mathbb Z^{+}$, we get
$$\begin{align}9k^2+12k-17=&9\times 17^2m^2+12\times 17m-17&\end{align}$$
Finally,
$$\begin{align}y=\frac{x^2-21}{17}
&=\frac{9\times 17^2m^2+12\times 17m-17}{17}\\
&=153m^2+12m-1\end{align}$$
One of the solution sets:
\begin{align}\color {gold}{\boxed {\color{black}{x=51m+2,  y=153m^2+12m-1.}}}\end{align}
Therefore, we have infinitely many integer solutions.
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