I know that (n-1)^2 | (n^k-1) means that (n-1)^2 *m= (n-1)(n^[k-1]+n^[k-2]+...+n+1). But how do I connect this with (n-1)*a=k? Thanks.
this is an implication "if and only if", <==> I'll do the part <==, then you try to do the part ==> prove that if n-1|k, then [tex](n-1)^2|n^k-1[/tex]: proof: first what is [tex]\varphi{([n-1]^2)}[/tex] ?? it is [tex](n-1)^{2-1}\cdot \varphi{(n-1)}=(n-1)\cdot \varphi{(n-1)}[/tex] now, by euler, as [tex]gcd(n-1,\ n)=1[/tex] then [tex]n^{\varphi{([n-1]^2)}}=n^{(n-1)\cdot \varphi{(n-1)}}\equiv\ 1\ mod\ (n-1)^2[/tex] and by Lagrange we know that either [tex]k=(n-1)\cdot \varphi{(n-1)}[/tex] or [tex]wk=(n-1)\cdot \varphi{(n-1)}[/tex] notice that [tex]\varphi{(n-1)}<n-1<k[/tex], and by hipothesis n-1|k EDIT: conclusion is: therefore [tex]n^k\equiv\ 1\ mod\ (n-1)^2[/tex]
Thank you! But at this moment we didnt study anything about Phi function. Do you know any other method?
oops, in fact it is (...)and by Lagrange we know that either [tex]k=(n-1)\cdot \varphi{(n-1)}[/tex] or [tex]k=(n-1)\cdot \varphi{(n-1)}w[/tex] popitar, try to rewrite and "play" with the equations you know related to it, factoring x^k-1 is a way, did your teacher give any similar problem with solutions?
Thank you so much, Al-Mahed! Factoring x^k - 1 is (x-1)(x^[k-1]+x^[k-2]+...+x+1), and I see that (x-1)^2 divides x^k-1means that (x-1) divides (x^[k-1]+x^[k-2]+...+x+1), but I don't see how I connect this with (x-1) divides k..
correct, k elements, now you must show some work on it, grab a coffe (if you like) and think about it hint: what means "x-1 divide k" in terms of euclidian form a=qb+r? how could you WRITE it down as an equation? and how to use it? you already saw by yourself you need to prove that x-1 divide x^[k-1]+x^[k-2]+...+x+1
(x-1) divides k means (x-1) * p = k for some p positive integer. Then I multiply it by (x-1) I get the following (x-1)^2*p=k*(x-1). Then I can say that x^k-1 = k*(x-1)? But (x-1)*z=x^[k-1]+x^[k-2]+...+x+1.. ahh
This is all I have (x-1)^2 | x^k - 1 <=> (x-1)|k (x-1)^2 * m = x^k -1 <=> (x-1)*z=k (x-1)^2 *m = (x-1)(x^[k-1]+x^[k-2]+...+x+1) <=> (x-1)*z=k (x-1)*m = x^[k-1]+x^[k-2]+...+x+1 <=> (x-1)*z=k
write x-1=a, then k=ma=m(x-1) for some integer m, since x-1| k, it yields x=a+1 x^[k-1]+x^[k-2]+...+x+1 = (a+1)^[k-1]+(a+1)^[k-2]+...+(a+1)+1 = a*S + k=(x-1)S+m(x-1), since you'll have k 1's, and the rest of it multiplied by a
Ok. So you said I replace (x-1) by a, and x=a+1 (x-1)^2 | x^k - 1 <=> (x-1)|k (x-1)^2 * m = x^k -1 <=> (x-1)*z=k (x-1)^2 *m = (x-1)(x^[k-1]+x^[k-2]+...+x+1) <=> (x-1)*z=k (x-1)*m = x^[k-1]+x^[k-2]+...+x+1 <=> (x-1)*z=k a*m=(a+1)^[k-1]+(a+1)^[k-2] + ...+ (a+1)+1 <=> a*z=k And from here what to do? Thanks!
We need to proofs: 1). If a|bc then a/d divides b, where d = gcd(a,b). 2). If a/d divides b then a|bc.
Popitar, I'm not a teacher, I don't even have a major in mathematics, I'm just a curious guy, so you have to check all this stuff with your teacher. Anyway, I think you are having trouble to understand the basic concepts of proof, but don't worry, you'll get it soon enough (it is quite confusing at the beginning). You have to understand the binomial theorem, its expansion, to see that you'll have all the expression being divisible by a=x-1. So if a valid substitution yields a multiple of x-1 you're done in your demonstration that x-1 divide the expression, as required. Now I think is a good idea turn out the internet and get pencil and paper... try to figure out some stuff by yourself now! Use a calculator to verify some ideas, you need to study more the basics, the euclidian algorithm is the most important thing right now. That's the stuff you must study now in order to advance. take care