Obtain the eight incongruent solutions of the linear congruence

[Math100]

Homework Statement

Obtain the eight incongruent solutions of the linear congruence $3x+4y\equiv 5\pmod {8}$.

Relevant Equations

None.

Consider the linear congruence $3x+4y\equiv 5\pmod {8}$.
Then $3x\equiv 5-4y\pmod {8}$.
Note that $gcd(3, 8)=1$ and $1\mid (5-4y)$.
Since $3^{-1}\equiv 3\pmod {8}$, it follows that $x\equiv 15-12y\pmod {8}\equiv 7+4y\pmod {8}$.
Thus ${(x, y)=(7+4y, y)\pmod {8}\mid 0\leq y\leq 7}$.
Therefore, $x\equiv 7, y\equiv 0; x\equiv 3, y\equiv 1; x\equiv 7, y\equiv 2; x\equiv 3, y\equiv 3;$
$x\equiv 7, y\equiv 4; x\equiv 3, y\equiv 5; x\equiv 7, y\equiv 6; x\equiv 3, y\equiv 7.$

 

[fresh_42]

Homework Statement:: Obtain the eight incongruent solutions of the linear congruence $3x+4y\equiv 5\pmod {8}$.
Relevant Equations:: None.

Consider the linear congruence $3x+4y\equiv 5\pmod {8}$.
Then $3x\equiv 5-4y\pmod {8}$.
Note that $gcd(3, 8)=1$ and $1\mid (5-4y)$.
Since $3^{-1}\equiv 3\pmod {8}$, it follows that $x\equiv 15-12y\pmod {8}\equiv 7+4y\pmod {8}$.
Thus ${(x, y)=(7+4y, y)\pmod {8}\mid 0\leq y\leq 7}$.
Therefore, $x\equiv 7, y\equiv 0; x\equiv 3, y\equiv 1; x\equiv 7, y\equiv 2; x\equiv 3, y\equiv 3;$
$x\equiv 7, y\equiv 4; x\equiv 3, y\equiv 5; x\equiv 7, y\equiv 6; x\equiv 3, y\equiv 7.$

That's right. The remainders modulo $8$ do not form a field because none of the even numbers has a multiplicative inverse. The odd remainders have such so that you could solve the equation.

 

[fresh_42]

And here is how the straight looks like. Of course, only the circled points do really count, so the green lines is a bit cheating.