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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.

Prove that a divides bc if and only if a/(gcd(a,b)) divides b.

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...

Can you explain this part :(a+1)^[k-1]+(a+1)^[k-2]+...+(a+1)+1 = a*S + k?

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

(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

I believe k elements.

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..

Thank you! But at this moment we didnt study anything about Phi function. Do you know any other method?

How do I check it?

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.

Let n>=2 and k>0 be integers. Prove that (n-1)^2 divides n^k -1 if and only if (n-1) divides k.

Homework Statement Prove that if V is a finite-dimensional vector space, then the space of all linear transformations on V is finite-dimensional, and find its dimension. Homework Equations The Attempt at a Solution

Thank you all!

Could you please help me with it? Please show me how this function is continuous at -1. Any help is appreciated..