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        riddles >> medium >> Sum The Exponents, Get Perfect Power?
        
(Message started by: K Sengupta on Jul 20^th, 2007, 11:28am)

Title: Sum The Exponents, Get Perfect Power?
Post by K Sengupta on Jul 20^th, 2007, 11:28am

Analytically determine, whether or not there exist any positive prime number x such that :
2^x + 5^x is a perfect power.

NOTE:

The definition of perfect power is given in:

http://mathworld.wolfram.com/PerfectPower.html

Title: Re: Sum The Exponents, Get Perfect Power?
Post by FiBsTeR on Jul 20^th, 2007, 1:04pm

(silly post removed)

Title: Re: Sum The Exponents, Get Perfect Power?
Post by SMQ on Jul 20^th, 2007, 1:30pm

Come again?

(pointing out of silly mistake removed -- no worries, mate, we all do it from time to time...)

--SMQ

Title: Re: Sum The Exponents, Get Perfect Power?
Post by FiBsTeR on Jul 20^th, 2007, 1:36pm

Oops.

Title: Re: Sum The Exponents, Get Perfect Power?
Post by SMQ on Jul 23^rd, 2007, 2:29pm

[hide]7 | 2^x + 5^x for all odd x, but 49 | 2^x + 5^x iff x is congruent to 7, 21 or 35 mod 42. Thus, for any odd prime x other than 7, 2^x + 5^x is multiple of 7 but not 49, and therefore not a perfect power.[/hide]

[hide]2² + 5² = 29 is not a perfect power. 2⁷ + 5⁷ = 78253 is not a perfect power.[/hide]

[hide]Therefore there is no x for which 2^x + 5^x is a perfect power.[/hide]

--SMQ

Title: Re: Sum The Exponents, Get Perfect Power?
Post by K Sengupta on Jul 24^th, 2007, 1:17am

on 07/23/07 at 14:29:29, SMQ wrote:

Well done, SMQ.

This was precisely the solution I was looking for.

However, your proof that :
(2^x+5^x)/7 = 0 (mod 7) iff x = 0 (Mod 7)
varies very slightly from my own methodology.

My proposed solution to the given problem is furnished hereunder as follows:

For x =2; 2^x + 5^x = 29, which is not a perfect power.
For x>=3, x must be odd, since x is prime.
Thus, (2^x + 5^x) is divisible by 7 for all x>=3, and accordingly, the perfect power, if it exists, must be a power of 7.

Hence substituting x = 2y+1, for positive integer y, we observe that:
(2^x + 5^x)/7
= (2^2y – 2^(2y-1)*5 + 2^(2y-2)*5^2 – 2^(2y-3)*(5^3) ……..- 2* 5^(2y-1) + 5^(2y))
= {2^(2y) + 2^(2y-1)*2 +2^(2y-2)*2^2 + ………+ 2*2^(2y-1) + 2(2y)}(Mod 7)
= ((2y+1)*2^(2y) (Mod 7)
= x* 2^(x-1)(Mod 7)

Since (2, 7) = 1, it now follows that (2^x + 5^x)/7 = 0 (Mod 7) iff x = 7t for some integer t.

However, we know that x is prime and accordingly, x= 7

Therefore, 2^x + 5^x may correspond to a perfect power only when x=7.

However, in reality, we observe that 2^7 + 5^7 = (7^2)*(19)*(83), and consequently:
2^x + 5^x can never correspond to a perfect power for prime x.