Solve for $x$ in the following congruence: $$x^2 − 7x ≡ 10 \pmod {11}$$
So this is what I thought we were supposed to do:
Rearrange the equation to get:
$$x^2 - 7x -10 ≡ 0 \pmod {11}$$
So I thought that I would find the roots of this equation to solve for $x$ but apparently that is not the approach.
Instead, the equation becomes:
$$x^2 - 7x + 12 ≡ 0 \pmod {11}$$
And therefore $x = 3$ and $4$.
But I am not exactly sure why the first equation became the second equation.
Any help?
On
Or just do a modular square completion: $$x^2-7x \equiv 10 \mbox{ mod } 11 $$ $$x^2-7x + 11x \equiv x^2 + 4x \equiv (x+2)^2 - 4 \equiv 10 \mbox{ mod } 11 $$
It follows:
$$(x+2)^2 - 14 \equiv (x+2)^2 - 14 -11 \equiv (x+2)^2 - 25 \equiv (x+2-5)(x+2+5) \equiv (x-3)(x+7) \equiv 0 \mbox{ mod } 11 $$
So, $$x \equiv 3 \mbox{ mod } 11 \mbox{ or } x \equiv -7 \equiv 4 \mbox{ mod } 11$$
$x^2 − 7x ≡ 10 $ (mod $11$) literally means that $10$ is the remainder of $x^2-7x \div 11$ with $x$ is an integer.
Put $x^2-7x=11k+10$ ($k$ is an integer), then $x^2-7x-10=11k$, so $x^2-7x-10$ is divisible by 11 or $x^2 − 7x -10≡ 0 $ (mod $11$).
Also, the equation actually has a generalized solution of $x=11k+3$ and $x=11k+4$ for all integers $k$, because
$(11k+3)^2-7(11k+3)-10=11(11k^2+3k-2)$ divisible by $11$ for all integers $k$, and
$(11k+4)^2-7(11k+4)-10=11(4k^2+k-2)$ divisible by $11$ for all integers $k$.
To prove this, the only way I know to put $x$ in 11 cases: $x=11k; x=11k+1;x=11k+2;x=11k+3$ and keep going until $x=11k+10$.
For example, with $x=11k+2$:
$(11k+2)^2-7(11k+2)-10=121k^2+44k+4-77k-14-10=44k^2-33k-20=11(4k^2-3k-2)+2$
not divisible by $11$ (because the remainder is $2$), so you can prove similarly for all other cases.