Sorry for the typescript,typescript; I felt more confident dointdoing it that way, but it should be fine to just remove the typings...
Sorry for the typescript, I felt more confident doint it that way, but it should be fine to just remove the typings...
Sorry for the typescript; I felt more confident doing it that way, but it should be fine to just remove the typings...
Let's define the inverse of mod m as invmod_m
(5) invmod_m(x mod m) = x
(6) invmod_m(x) mod m = x
(57) invmodinvmod_m((b − a) mod m, m) == invmodinvmod_m(0, m)
from (5) we can simplify the left side
(68) b - a == invmodinvmod_m(0, m)
(79) b == a + invmodinvmod_m(0, m)
(810) b mod m == (a + invmodinvmod_m(0, m)) mod m
(911) b mod m == a mod m + invmodinvmod_m(0, m) mod m
from (6) we see that invmodinvmod_m(0, m) mod m is clearly zero so we have
(1012) (b mod m) == (a mod m)
(5) invmod((b − a) mod m, m) == invmod(0, m)
(6) b - a == invmod(0, m)
(7) b == a + invmod(0, m)
(8) b mod m == (a + invmod(0, m)) mod m
(9) b mod m == a mod m + invmod(0, m) mod m
invmod(0, m) mod m is clearly zero so we have
(10) (b mod m) == (a mod m)
Let's define the inverse of mod m as invmod_m
(5) invmod_m(x mod m) = x
(6) invmod_m(x) mod m = x
(7) invmod_m((b − a) mod m) == invmod_m(0)
from (5) we can simplify the left side
(8) b - a == invmod_m(0)
(9) b == a + invmod_m(0)
(10) b mod m == (a + invmod_m(0)) mod m
(11) b mod m == a mod m + invmod_m(0) mod m
from (6) we see that invmod_m(0) mod m is zero so we have
(12) (b mod m) == (a mod m)
by applying the mod inverse we get infinitely many equations, but that' ok, they look all the same and we can parametrizedeal with an integer kthem all at once
(5) invmod((b − a) mod m, m) == invmod(0, m)
(6) b - a == k*m + invmod(0, m)
(7) b == a + k*minvmod(0, m)
(8) b mod m == (a + k*minvmod(0, m)) mod m
(9) b mod m == a mod m + invmod(k*m0, m) mod m
invmod(k*m0, m) mod m is clearly zero so we have
by applying the mod inverse we get infinitely many equations, but that' ok, they look all the same and we can parametrize with an integer k
(5) invmod((b − a) mod m, m) == invmod(0, m)
(6) b - a == k*m + 0
(7) b == a + k*m
(8) b mod m == (a + k*m) mod m
(9) b mod m == a mod m + (k*m) mod m
(k*m) mod m is clearly zero so we have
by applying the mod inverse we get infinitely many equations, but that' ok, we can deal with them all at once
(5) invmod((b − a) mod m, m) == invmod(0, m)
(6) b - a == invmod(0, m)
(7) b == a + invmod(0, m)
(8) b mod m == (a + invmod(0, m)) mod m
(9) b mod m == a mod m + invmod(0, m) mod m
invmod(0, m) mod m is clearly zero so we have