## 2^k+1 never divisible by u(8x-1) in integers

It seems that there exists no integer k such that 2^k+1 is divisible by a positive integer n, if and only if n is of the form u(8x-1) (where u and x are also both positive integers).

How could this be proved/disproved?

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 An even integer is defined by 2n for some integer n, and an odd integer is defined as 2m+1 for some integer m. 2^k+1 =2*(2*2*...*2)+1 where (2*2*...*2) is an integer by closure properties therefore 2^k+1 is odd by definition of odd u(8x-1) u,x arbitrary positive integers =2A(8*(2B)-1) A,B arbitrary integers =16AB-2A =2(8AB-A) (8AB-A) is an integer by closure properties Suppose that 2(8AB-A) divides 2^k+1. Then 2^k+1=2(8AB-A)*P for some integer P =2*(ABP-AP) => even Which is a contradiction, an odd cannot equal an even. But, the claim is that an there does not exist an integer n if and only if n=u(8x-1) ... Counter example, let n=c(2d-1) c, and d positive integers =2r*(2*(2s)-1) r,s arbitrary integers =2r*(4s-1) =2*(4rs-r) If 2*(4rs-r) divides 2^k+1, then 2^k+1=2*(4rs-r)*t for some integer t =2*(4rst-rt) => even even cannot equal an odd So there is at least one integer other than u(8x-1) such that it does not divide 2^k+1 for any k So the original assumption is false I don't know for sure if this what your looking for or completely sure it's correct. I'm new to number theory.
 your argument simply put is that 2^k+1 wouldn't be divisible by any even number, indeed this is correct. I forgot to specify that n here must also be odd.

## 2^k+1 never divisible by u(8x-1) in integers

 Quote by jreelawg An even integer is defined by 2n for some integer n, and an odd integer is defined as 2m+1 for some integer m. 2^k+1 =2*(2*2*...*2)+1 where (2*2*...*2) is an integer by closure properties therefore 2^k+1 is odd by definition of odd u(8x-1) u,x arbitrary positive integers =2A(8*(2B)-1) A,B arbitrary integers

Why have you decided that both u, x are even numbers? This is not given in the OP and, of course, then these are not "arbitrary integeres".

 =16AB-2A

This is incorrect: $\,\,2A\left(8(2B)-1\right)=32AB-2A$

 =2(8AB-A) (8AB-A) is an integer by closure properties Suppose that 2(8AB-A) divides 2^k+1. Then 2^k+1=2(8AB-A)*P for some integer P =2*(ABP-AP) => even Which is a contradiction, an odd cannot equal an even. But, the claim is that an there does not exist an integer n if and only if n=u(8x-1) ... Counter example, let n=c(2d-1) c, and d positive integers =2r*(2*(2s)-1) r,s arbitrary integers =2r*(4s-1) =2*(4rs-r) If 2*(4rs-r) divides 2^k+1, then 2^k+1=2*(4rs-r)*t for some integer t =2*(4rst-rt) => even even cannot equal an odd So there is at least one integer other than u(8x-1) such that it does not divide 2^k+1 for any k So the original assumption is false I don't know for sure if this what your looking for or completely sure it's correct. I'm new to number theory.
DonAntonio

 I'm pretty sure the if part of this conjecture involves proving that 8gx - g - x cannot be a power of two. 2^k+1/u(8x-1) = m; in integers Then 2^k = um(8x-1) - 1 for this to possibly be true (um) must be of form 8g-1 because subtracting a number that isn't divisible by 8 from a number that is divisible by 8 won't yield a result divisible by 8 and you need one divisible by 8 because it has to be a power of two and it isn't 2 or 4 because um(8x-1) cannot be 3 or 5 as x is an integer (the smallest possible value of um(8x-1) is 7). (8g-1)(8x-1) - 1 = 2^k and k is greater than or equal to 3 (64gx -8g - 8x +1) - 1 = 2^k 8gx - g - x = 2^(k-3) ...this is where i'm at. Additional logic: the smallest possible value of 8gx - g - x is 6 therefore analogously the whole thing must be divisible by eight and for that to be true g+x must also be divisible by eight so g+x can be expressed as 8t g=8t-x 8x(8t-x) - 8t = 2^(k-3) 64tx - 8x^2 - 8t = 2^(k-3) 8tx - x^2 - t = 2^(k-6) 8(g+x)x - x^2 - x - g = 2^(k-6) 8g + 8x^2 - x^2 - x - g = 2^(k-6) 7g + 7x^2 - x = 2^(k-6) ...

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 Quote by nkpstn It seems that there exists no integer k such that 2^k+1 is divisible by a positive integer n, if and only if n is of the form u(8x-1) (where u and x are also both positive integers).
Your wording strikes me as a little odd (the "and only if" bit), so to make sure we're on the same page I'll paraphrase it:
If n is a factor of 2^k+1 then n is not congruent to 7 mod 8.

I believe this is a special case of a much more general circumstance that appears to be true:
If (x,y)=1 [i.e. x and y are coprime] then
a) No factor of x$^{2}$+y$^{2}$ is congruent to 3 or 7 mod 8
b) No factor of 2x$^{2}$+y$^{2}$ is congruent to 5 or 7 mod 8
(Your conjecture follows from (a) for k even and (b) for k odd.)

I'll sketch a proof of (a):
x$^{2}$+y$^{2}$ can be factorised as x+iy, x-iy.
Since x and y coprime, neither of these complex factors has any real factors greater than 1.
If n is a real factor of x$^{2}$+y$^{2}$, it must be factorisable into complex factors of x+iy, x-iy.
If these are a+ib, c+id respectively, ad + bc = 0.
Since (x,y)=1, (a,b)=1 and (c,d)=1.
So every real factor of a is also a real factor of c and vice versa.
Hence a = +/-c, and b = +/-d.
Thus, w.l.o.g., a = c, d = -b.
So n = a$^{2}$+b$^{2}$. Since each square is individually 0, 1 or 4 modulo 8, the sum must be 0, 1, 2, 4 or 5 modulo 8.

 ah I see. Although I do believe referring to complex numbers is over complicating it. I've also found that the only if part of my conjecture is untrue with a counterexample of 73. (I checked all the way to 71 before the first post) I guess there is no more purpose for this thread. Though I still am wondering how to prove that 8gx - g - x can't be a power of 2... eh I'll probably figure it out on my own later.