<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] $x^2-21=17y$ has integer solutions precisely when $x^2-4=17(y+1)$ has integer solutions, which occurs precisely when it's possible to solve $x^2 \equiv 4 \pmod{17}$.  This has solutions, of course, when $x \equiv \pm 2 \pmod{17}$.
Checking:  $x =2, x^2-21=-17=17(-1); x=15, x^2-21=204=17(12)$.
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