- #1
popitar
- 17
- 0
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.
al-mahed said:you may want to check [tex]\varphi{([n-1]^2)}[/tex]
popitar said:How do I check it?
al-mahed said:(...)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]
popitar said: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..
popitar said:I believe k elements.
popitar said:Can you explain this part :(a+1)^[k-1]+(a+1)^[k-2]+...+(a+1)+1 = a*S + k?
It means that the expression n^k - 1 is divisible by (n-1)^2 without leaving any remainder.
This expression is commonly used in number theory and has various applications in cryptography, coding theory, and other mathematical fields.
There are multiple ways to prove this statement, including using mathematical induction, polynomial division, or the binomial theorem.
Yes, this statement holds true for all positive integer values of n and k.
Sure, for n=3 and k=4, (n-1)^2 = (3-1)^2 = 4, and n^k - 1 = 81-1 = 80. Since 80 is divisible by 4 without any remainder, we can say that (n-1)^2 divides n^k -1 in this case.