Theorem I (Fermat) : for prime p  if gcd(a,p) = 1 then a^(p-1) - 1 is divisible by p. 

1.0   3^x + 5^x - 2 = 11k for some integers x and k.

Proof: We rewrite 1.0 as 

1.1    3^x -1 + 5^x - 1 = 11k

For x = 11, from  Theorem I, we have 11 | 3^10 - 1 and 11 | 5^10 -1. Then 10z is a solution of 1.1. Also,  3^10 = 9^5  => 11 | 9^5 - 1. Similarly, 11 divides 25^5-1. Thus, 5z is a general solution for all integer z. Then  (3^z)^5 -1 = 11s and (5^z)^5 - 1=11t or  3^5z - 1 = 11s and 5^5z-1 = 11t. Now by subtracting 11 from both sides of  1.1, we have 3^x + 5^x -13 = 11(k-1). We now can test 3^(5z-1) -4 and  5^(5z-1)-9.  Multiplying respectively by 3 and 5 we get 3^5z-1 + 5^5z-1 -55 =11k. So x=5z and x=5z-1 are general solution to 1.0.

2.0   3^x + 5^x - 1 = 11k for some integers x and k.

Test

2.1   3^5z-1 + 5^5z - 1 +1 = 11k.

 Multiply terms respectively by 9 and 25, we get  3^(5z+2) - 9 + 5^(5z+2) - 25 + 1 = 11k = 3^(5z+2) + 5^(5z+2) -34+1. So x=5x+2 is a general solution to 2.0.

3.0    3^x + 5^x - 8 = 11k for some integers x and k.

From 1.1 we have  3^5x -1 + 5^x - 1 = 11k. Multiplying by 3 and 5 we get  3^(5x+1) -3 + 5^(5x+1)-5  = 3^(5x+1) + 5^(5x+1) - 8 = 11k. Thus, x=5x+1 is a general solution 3.0.

4.0   3^x + 5^x + 2 = 11k

Test

4.1   3^(5z+3) + 5^(5z+3) +2 = 11k

subtracting 154 from both sides

3^(5z+3) - 27 + 5^(5x+3) - 125 = 11k-11*14 = 11t  

27(3^5z-1) + 125(5^5z-1) = 11t

From 1.1, 3^5z-1 and 5^5z-1 are divisible by 11. Therefore, x = 5z+3 is a general solution to 4.0.

If we subtract 11 from both sides of  4.1 we have 

3^(5z+3) + 5^(5z+3) - 9 = 11(k-1).  we have x=5z+3 as a general solution to

5.0   3^x+5^x - 9 = 11k.     

Cino Hilliard
Jun 2003