Maths -- Factorization

1. 10000...00001 (2003 0s)
= 10^24 + 1
= (10^8)^3 + 1^3
= (10^8 + 1)(10^16 - 10^8 + 1)
So the number is composite
2. (a - b)^2 + (b - c)^2 + (c - a)^2
= a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2
= 2(a^2 + b^2 + c^2) - 2(ab + bc + ca)
By Cauchy inequality,
(ab + bc + ca)^2 <= (a^2 + b^2 + c^2)(b^2 + c^2 + a^2)
(ab + bc + ca)^2 <= 2008^2
-2008 <= ab + bc + ca <= 2008
Therefore maximum value for (a - b)^2 + (b - c)^2 + (c - a)^2
= 2(2008) - 2(-2008)
= 8032
3. The general term is 1 + 4 + 4^2 + ... + 4^n
= [4^(n+1) - 1] / (4 - 1)
= [2^(2n+2) - 1] / 3
= [2^(n+1) + 1][2^(n+1) - 1] / 3
when n = 0, the number is (3)(1)/3 = 1 not prime
when n = 1, the number is (5)(3)/3 = 3 is prime
when n >=2, the number is the product of [2^(n+1) + 1] and [2^(n+1) - 1] divided by 3. One of these bracketed terms must be divisible by 3. As 2^(n+1) >= 8 for n >= 2, the quotient is bigger than one. Therefore the number must be the product of two integers > 1.
There is only one prime member in S which is 5.

2009-10-26 22:39:22 補充：
Correction (the original solution mistakenly took the values as binary numers):
3. The general term is 1 + 100 + 100^2 + ... + 100^n
= [10^(n+1) - 1][10^(n+1) + 1] / 99
n = 0 => Value = 1
n = 1 => Value = 101 is prime

2009-10-26 22:39:50 補充：
when n >=2, the number is the product of [10^(n+1) + 1] and [10^(n+1) - 1] divided by 99. As 10^(n+1) >= 1000 for n >= 2, so both bracketed terms are much greater than 99, and each term may be divisible by 3, 9, 11, 33, or 99.

2009-10-26 22:40:23 補充：
In all cases, the quotients after divisions are always > 1 so the value is a product of 2 integers => the value is composite.
There is only one prime member in S which is 101.