Please help me to Show that i. a = b(mod m)=> (a,m) = (b,m) ii. a = b(mod m)=> a^n = b^n (mod m) iii. ak=bk(mod m)=> b{mod[m/(k,m)]}
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Originally Posted by dhammikai Please help me to Show that i. a = b(mod m)=> (a,m) = (b,m) ii. a = b(mod m)=> a^n = b^n (mod m) iii. ak=bk(mod m)=> b{mod[m/(k,m)]} Can we see where are you stuck here exactly plz?
I am trying to solve this form following please see, but i am stuck with how I implement it
Originally Posted by dhammikai I am trying to solve this form following please see, but i am stuck with how I implement it Plz avoid images as much as possible. At times you just don't feel like clicking and opening them. Why don't you start. So. let's see some working for (1). I will give you a hint if x|y and y|x => x=y.
(i) is talking about the Euclidean Algorithem. (ii) and (iii) are talking about the property of congruence concernning the power and division operation. So just follow the definition and the hint, you will get closer to the conclusion.
1.) a = b(mod m)=> (a,m) = (b,m) "Why don't you start. So. let's see some working for (1). I will give you a hint if x|y and y|x => x=y." then, If x = y (mod m) then y = x+sm for some s, is true?
Originally Posted by dhammikai 1.) a = b(mod m)=> (a,m) = (b,m) "Why don't you start. So. let's see some working for (1). I will give you a hint if x|y and y|x => x=y." then, If x = y (mod m) then y = x+sm for some s, is true? yes, more exactly (x-y)=sm but you have just stated the definition of the congruence relation
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