Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: cotpi 46 - Raised to the same power
Replies: 12   Last Post: Apr 10, 2012 4:26 PM

 Messages: [ Previous | Next ]
 GG Posts: 3 Registered: 4/9/12
Re: cotpi 46 - Raised to the same power
Posted: Apr 9, 2012 7:22 AM

On Monday, 9 April 2012 10:47:56 UTC+5:30, quasi wrote:
> On Sun, 08 Apr 2012 21:48:26 +0530, cotpi
> wrote:
>

> >Two given positive integers are raised to the same power and
> >added. The sum is a power of 2. What is the maximum possible
> >difference between the two given integers?

>
> Zero (assuming that by "power" you mean a power greater than 1).
>
> Proposition:
>
> If x^n + y^n = 2^k where x,y,n,k are positive integers and
> n > 1, then x = y (hence x - y = 0).
>
> Proof:
>
> If k = 1, x^n + y^n = 2 => x^n = y^n = 1 => x = y = 1.
>
> If k = 2, x^n + y^n = 4 => x^n = y^n = 2 => n = 1,
>
> Assuming there is a counterexample, choose one with k as small
> as possible.
>
> Necessarily, k >= 3.
>
> x^n + y^n = 2^k
>
> => x,y have the same parity
>
> But if x,y are both even, we get
>
> (x/2)^n + (y/2)^n = 2^(k-n)
>
> Since x/2, y/2 are positive integers, k-n must also be a
> positive integer, but then the above equation would give a
> counterexample with a smaller power of 2, contrary to choice
> of k.
>
> Thus, x,y are both odd.
>
> Suppose first that n is even.
>
> Since k >= 3, we have
>
> x^n + y^n = 0 mod 8
>
> which is impossible since x^n and y^n would be odd squares,
> and odd squares are congruent to 1 mod 8.
>
> Hence n is odd.
>
> Factoring x^n + y^n, we get
>
> a*b = 2^k
>
> where
>
> a = x + y
>
> b = x^(n-1) + x^(n-2)*y + ... + x*y^(n-2) + y^(n-1)
>
> Note that b is the sum of n terms, each of which is odd,
> hence, since n is odd, b is odd.
>
> But b is a factor of 2^k, hence b = 1.
>
> But then
>
> x^n + y^n = a*b = a = x + y
>
> Now x,y are not both 1 (else k = 2), hence, without loss of
> generality, assume y > 1.
>
> But then x <= x^n and y < y^n, which yields x + y < x^n + y^n,
>
> Hence there can't be a counterexample, which completes the
> proof.
>
> quasi

Clarifications with your proof:
1.you said that when x,y are even

(x/2)^n + (y/2)^n = 2^(k-n)

here how do we know that (k-n) is a positive integer.

2.Further down you say when x,y are odd

x^n + y^n is congruent to 1 (mod 8)

Why should it be so , when (x^n + y^n) is even. [since x^n and y^n are both odd].

Date Subject Author
4/8/12 cotpi
4/8/12 Prai Jei
4/8/12 cotpi