This is true for n = 1, because 3^(4+2) + 5^(2+1) = 854 = 14 * 61.

Inductive Step:Assuming that this is true for n = k,3^(4(k+1) + 2) + 5^(2(k+1) + 1)= 3^(4k+6) + 5^(2k+3)= 3^4 * 3^(4k+2) + 5^2 * 5^(2k+1)= 81 * 3^(4k+2) + 25 * 5^(2k+1)= (25 + 56) * 3^(4k+2) + 25 * 5^(2k+1)= 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2).

By inductive hypothesis, 3^(4k+2) + 5^(2k+1) is divisible by 14.Clearly, 56 * 3^(4k+2) = 14 * 4 * 3^(4k+2) is divisible by 14.

Therefore, 3^(4(k+1) + 2) + 5^(2(k+1) + 1) = 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2)is divisible by 14, completing the inductive step.-------------------2) You meant to write x^n - 1 is divisible by x - 1 for x ≠ 1.

Base Case:This is true for n = 1, since x - 1 is clearly divisible by x - 1.

Inductive Step:Assuming that this is true for n = k,x^(k+1) - 1= (x^(k+1) - x) + (x - 1)= x(x^k - 1) + (x - 1).

By inductive hypothesis, x^k - 1 is divisible by x - 1.Clearly, x - 1 is divisible by x - 1.

So, x^(k+1) - 1 = x(x^k - 1) + (x - 1) is divisible by x -