# 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?

$$x^2-21=17y$$ 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: $$x =2, x^2-21=-17=17(-1); x=15, x^2-21=204=17(12)$$.

$$x=2, y=-1$$ is an integer solution.

$$x^2-21=17y$$

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.