Algebra 挑戰題 2

Prime nos., apart from 2, are all odd nos. And in this case, z cannot be equal to 2 as x^y must not be equal to 1.
∴z is odd, x^y is even, and , as for "odd to the power odd" results in odd nos., which does not match the result that x^y should be odd. Also odd square also results in odd. So, there exists only one case where x=2
Consider the situation 2^n+1 is a multiple of 3 for all odd n
n=1
LHS=3=3x1
P(1) is true
Assume that P(k) is true where k is an odd integer.
i.e. 2^k+1=3M is true, where M is a odd no.
For n=k+2
2^(k+2)+1
=4(2^k)+1
=4(3M-1)+1
=12M-3
=3(4M-1), which is divisible by 3
∴P(k+1) is true
By MI, the statement that 2^n+1 is divisible by 3 is true for all odd nos. n
∴There exists contradiction for the situation as from the above MI, it is proved that z is a multiple of 3
∴There is no soultion

It is easy to argue that x = 2, which means that 2^y +1 = z is prime.

But this implies that y is a power of 2, contradicting to y not equal to x = 2. (So, we don't even need y to be odd prime in order to derive a contradiction. We just need y contains prime factor other than 2)

2008-10-19 17:08:28 補充：
To finish the proof, let us show that whenever 2^n + 1 is prime, then n must be a power of 2 (including n =1)

If n = ab and b is odd, then
2^n + 1 ≡ (2^a)^b + 1 ≡ (−1)^b + 1 ≡ 0 (mod (2^a + 1)). This contradicts to 2^n + 1 being prime. Thus, all factos of n are even, i.e., n is a power of 2.

2008-10-19 17:10:20 補充：
The proof by hawk_wing is simply taking a = 1 in the above proof.

除左x=2 , y=2 , z=5 , xyz=20
但x, y and z are distinct prime integers
即是三個不同的數?

(1)All prime numbers are odd numbers. So for z to be a prime number, x^y must be even.
(2) For x to be a prime number and at the same time x^y is even, x can only be 2 since this is the only even prime number.(If the unit digit of x is odd, x^y must be odd, so x^y +1 must be even and z can never be a prime number.)
(3) Since y is also a prime number, y must be odd. So we put 2^y +1(= z) as [ 2^(2n-1) +1] where n can be any positive integer.
(4)However, it can be proved by mathematical induction that 2^(2n-1) +1 is divisible by 3 for all positive integer of n, that means z is divisible by 3 for any value of y and never be a prime number.
So in summary, you can never find 3 prime numbers that can satisfy x^y +1 = z. So no solution for xyz.