How do you solve $x^2−4\equiv 0 \mod 7$?

Question

hint: how to solve $ab\equiv 0\pmod 7$ ? – zwim Dec 24 '20 at 11:31 — zwim, Dec 24 '20 at 11:31
i have done it thx alot – Mohamad Abo allo Dec 24 '20 at 11:37 — Mohamad Abo allo, Dec 24 '20 at 11:37
$x^2\equiv4\iff x\equiv\pm2$ – J. W. Tanner Dec 24 '20 at 13:55 — J. W. Tanner, Dec 24 '20 at 13:55

score 2 · Accepted Answer · edited Dec 24 '20 at 12:00

2

The most efficient method here is that $x^2-4=(x-2)(x+2)$, thus it implies that $7$ divides $(x-2)(x+2)$ but because $7$ is a prime, it must divide either $x+2$ or $x-2$, therefore the solutions are $x=±2$ (mod $7$)

edited Dec 24 '20 at 12:00

The 2nd

856

answered Dec 24 '20 at 11:37

Aryan

1,508

true and also will have x=±5 – Mohamad Abo allo Dec 24 '20 at 11:39
1

Those are the same mod7 – Aryan Dec 24 '20 at 11:41
yea and that what we will need , thx alot – Mohamad Abo allo Dec 24 '20 at 11:42
what if we have x^2 + 3x ≡ 4 ( mod 5 )? we can do that x^2 + 3x ≡ 4 is equal to (x+4)(x-1) ≡ 0 ( mod 5 ) ? if its right how i can continue that ? – Mohamad Abo allo Dec 24 '20 at 11:48
Yeah its completely correct. Now because 5 is a prime the only possibilities are that 5 divides one of those two – Aryan Dec 24 '20 at 11:51

How do you solve $x^2−4\equiv 0 \mod 7$?

1 Answers1