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, > contradiction. > > 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, > contradiction. > > 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].