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nowimpsbball
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Homework Statement
Prove that x^100 = 1 for all x in U(1000)
Homework Equations
The Attempt at a Solution
U(1000) is all numbers relatively prime to 1000. And that is as far as I have gotten.
Thanks
nowimpsbball said:Well that is the problem how it is stated...but the idiot I am keeps forgetting to mention all the details of U(n)...it is the group of units modulo n (that is the set of integers less than n and relatively prime to n under multiplication modulo n). But what you say is what I want.
I for example take 3^100=1mod(1000) , 1000 should divide (3^100 - 1) but it does not, you end up with a decimal...so I can't even see what the problem looks like when you plug in a number to x
No, U(1000) refers to the set of integers that are relatively prime to 1000, meaning they do not share any common factors with 1000 except for 1. This set includes numbers such as 1, 3, 7, 9, 11, 13, etc.
We can prove this by using Euler's theorem, which states that if a and m are relatively prime integers, then a^phi(m) is congruent to 1 mod m, where phi(m) is the Euler totient function. In this case, phi(1000) = 400, so x^400 is congruent to 1 mod 1000 for all x in U(1000). Therefore, x^100 = 1 mod 1000 for all x in U(1000).
Yes, we can use induction to prove this statement. We can start with x = 1, and show that 1^100 = 1 for all x in U(1000). Then, we can assume that x^100 = 1 for some x in U(1000), and use this assumption to prove that (x+1)^100 = 1 for x+1 in U(1000). This will complete the inductive step, and we can conclude that x^100 = 1 for all x in U(1000).
Yes, this statement can be extended to other values of x and m, as long as x and m are relatively prime. The proof will follow the same logic using Euler's theorem and the Euler totient function.
This statement has applications in cryptography, particularly in the RSA algorithm. It is also used in number theory and modular arithmetic. Additionally, it can be used to simplify calculations in certain mathematical problems.