I have this congruence $ 2x \equiv a \mod 22 $ , so I used the chinese theorem and what I've got is $ a \equiv 0 \mod 2 $ and $ x \equiv 0 \mod 11 $ . I did it well? If yes, how should I proceed?
How do I proceed with this congruence?
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linear-algebra
discrete-mathematics
congruences
2 Answers
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I'm assuming that $a$ is an (unknown) parameter. The first part ($a \equiv 0 \pmod 2$) is correct, it means that the unknown parameter must be even in order for the equation to have any solution. The second one is not correct, though. You get
$$2x \equiv a \pmod {11}$$
And now you can divide both sides by 2 (because 2 has no common divisor with the modulus 11) and get
$$x \equiv \frac{a}{2} \pmod {11}$$
We know that you can divide $a$ by 2 because of the first property.
So the result is that you only get a solution if $a$ is even and (in that case) x has a single solution mod 11.
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First of all we have to be sure that $a$ is an even number in order to have a solution, otherwise there is no solution, as $2$ doesn't have a inverse modulo $22$. Then you just divide by $2$, but also you have to divide the modulo, as $(22,2) \not = 1$ and you will get that $x \equiv \frac {a}{2} \pmod{11}$
Actually this method is the Chinese Remainder Theorem in disguise. Assume that there is solutions s.t. $2x \equiv a \pmod{22}$, then we have $a \equiv 0 \pmod{2}$ and $x \equiv a2^{-1} \pmod {11}$. As previously we must have $a$ is even and then $x \equiv \frac a2 \pmod {11}$