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## Main Question or Discussion Point

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.

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- #1

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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.

- #2

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you may want to check [tex]\varphi{([n-1]^2)}[/tex]

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- #4

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How do I check it?you may want to check [tex]\varphi{([n-1]^2)}[/tex]

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this is an implication "if and only if", <==>How do I check it?

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]

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- #7

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oops, in fact it is(...)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]

(...)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?

- #8

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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..

- #9

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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..

how many elements you have in x^[k-1]+x^[k-2]+...+x+1 ?

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I believe k elements.

- #11

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I believe k elements.

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

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- #13

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(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

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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

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

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(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!

- #17

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

- #18

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1). If a|bc then a/d divides b, where d = gcd(a,b).

2). If a/d divides b then a|bc.

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

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

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