About the solvability of a system of congruence

Asked Dec 13 '20 at 18:22

Active Dec 14 '20 at 09:48

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I am asking if there is any method to decide if the following system of congruence is solvable or not:

$x_{1}$≡$a_{1}$ $mod (y_{1})$

$x_{2}$≡$a_{2}$$mod (y_{2})$

$x_{3}$≡$(a_{1}+a_{2})$$mod (y_{3})$

Here $y_{1},y_{2},y_{3}$ are not necessarily coprime integers

asked Dec 13 '20 at 18:22

Safwane

By the CRT solvability criterion it is solvable $\iff (y_1,y_2)\mid a_1!-!a_2,,$ $,(y_1,y_3)\mid a_2,,$ $,(y_2,y_3)\mid a_1$ $\qquad$ – Bill Dubuque Dec 13 '20 at 18:56
@BillDubuque: We have not the same $x$ as in the CRT. We have three distinct integers. – Safwane Dec 14 '20 at 09:26
Oops, missed that. Now reopened. Which symbols are variables (vs. constants or parameter), i.e. what are you solving for? – Bill Dubuque Dec 14 '20 at 09:48
@BillDubuque: the variables are x1,x2,x3. – Safwane Dec 14 '20 at 09:52
So we are given all $a_i$ and $y_i$ and you seek to find $x_i$ satisfying those congruences? If so they are independent and you can trivially solve each separately, e.g. $,(x_1,x_2,x_3) = (a_1,a_2,a_1+a_2),$ is a solution. – Bill Dubuque Dec 14 '20 at 09:59
@BillDubuque: They are not independent as equation 3 – Safwane Dec 14 '20 at 10:02
Independent means that a variable $x_i$ doesn't appear in more than one equation so you can solve for $x_i$ from the lone equation it occurs in. – Bill Dubuque Dec 14 '20 at 10:03
@BillDubuque: Can we find sufficient and necessary condition for the solvability. – Safwane Dec 14 '20 at 10:04
It's always solvable. I gave one simple solution above. – Bill Dubuque Dec 14 '20 at 10:05

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