Has the equation $x^2-21 = 17y$ integer solutions?


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?


3 Answers 3


$x^2-21=17y$ 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.



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

$$\begin{align}9k^2+12k-17=&9\times 17^2m^2+12\times 17m-17&\end{align}$$


$$\begin{align}y=\frac{x^2-21}{17} &=\frac{9\times 17^2m^2+12\times 17m-17}{17}\\ &=153m^2+12m-1\end{align}$$

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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