1
$\begingroup$

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?

$\endgroup$

3 Answers 3

2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.