Is anyone on /sci/ actually able to do theoretical math (e.g. number theory). Or do they just shit on math because they are bad at it?
Find all natural numbers n such that the number [math]2^n−1[/math] is divisible by 7.
Prove that for all natural numbers n the number [math]2^n+1[/math] is not divisible by 7.
>>7681050
Yes these problems are relatively easy. I picked them since you don't have to have too much specialized knowledge or training.
The first one is just all n equivalent to 0 modulo 3. The value of 2^n - 1 modulo 7 is periodic in n with period 3, and the values that occur are 1, 3, and 7 (or 0, if you prefer). It's easy to show inductively that this pattern continues by noting that to get from 2^n - 1 to 2^{n+1} - 1, you multiply by 2 and then add 1.
The second one can be done in essentially the same way, but the values modulo 7 which occur are 3, 5, and 2, so none of those numbers are multiples of 7.
>>7681050
Did the goat die?
2^3 = 1 modulo 7
hence 2^n = 2^(n-3) mod 7
so in both cases it suffices to check the few first cases
we conclude that 7|2^n - 1 iff 3|n
and that 7 never divides 2^n+1, since it doesnt divide any of 2^0+1, 2^1+1, 2^2+1
next problem pls
>>7681093
>>7681105
>Doing the kid's homework
>>7681050
encore toi...
>>7681105
Find a, b [math]\in {\bf Z}[/math] such that
ab(a+b) is NOT divisible by 7 but [math](a+b)^7 - a^7 - b^7[/math] is divisible by [math]7^6[/math]
>>7681173
oops [math]7^7[/math], not [math]7^6[/math]
Thought about taking a number theory course,but i don't have the prereqs.
The most upper level class I'll be taking is ode's . At my school you need some intro to proofs class.
I ain't bout that life.
Number theory was never my favorite subject. So much of it is not done for the sake of understanding, but just create increasingly-esoteric answers to increasingly-irrelevant questions.
I get that conclusions in number theory often prove to be extremely relevant and necessary for fields and proofs formed centuries later. But, you know, just not my cup of tea.
>>7681050
The second one we can easily prove with strong induction. If 2^n + 1 is not a multiple of 7, and because 8 is not a multiple of 7, we have it that
8*(2^n + 1) = 2^(n+3) + 8 is not a multiple of 7. But if 2^(n+3) + 8 is not a multiple of 7, then 2^(n+3) + 1 cannot be a multiple of 7. So, if we prove that 3, 5 and 9 are not multiples of 7, then we can use strong induction to show that 2^n + 1 is not a multiple of 7 for any positive integer n. The first problem is easy, just take 2^0 - 1, prove it is a multiple of 7, then prove 2^1 - 1 is not a multiple of 7, and prove 2^2 - 1 is not a multiple of 7. With strong induction, you can produce all natural numbers n such that 2^n - 1 is a multiple of 7, and those numbers n are the multiples of 3.
>>7681227
The first statement is true because 7 is a prime number. If 8*(2^n + 1) is divisible by 7, then the face that 7 is prime means that 7 must divide 8 or divide 2^n + 1 (or both). However, according to our inductive hypothesis, and because 8 = 7*(1) + 1, we know that 8*(2^n + 1) is not a multiple of 7.
>>7681227
Nevermind about the strong induction in the last part, just use regular induction three times.
>>7681173
we do some multiplying and we get
[eqn](a+b)^7 - a^7 - b^7 = 7ab(a+b)(a^2 + ab + b^2)^2 [/eqn]
so we need to find a, b such that [math]7^3 | a^2 + ab + b^2 [/math]
this can be transformed into
[eqn] \frac{2a+b}{b}^2 \equiv -3 \pmod{7^3} [/eqn]
so we first solve x^2 = -3 mod 7^3, this is done by first solving mod 7, then mod 7^2, then mod 7^3
i arrived at solution x = +- 37 mod 7^3
plug this back, and a surprisingly small solution is possible: a = 18, b = 1
fast check: 18^2 + 18 + 1 = 343 = 7^3
am i
>able to do theoretical math (e.g. number theory)
already? if not, pls post more
>>7681050
global rule 13
