5^n + 8^(n+1) + 13^(n + 2) = (-1)^n + 2^(n+1) + 1 (mod 6)
Case 1: n is odd
2^(n + 1) = 0 (mod 6)
No values of n will satisfy this as we will need to get a factor of 3 in order to make this expression divisible by 6.
Case 2: n is even
2^(n+1) + 2 = 0 (mod 6)
2^(n+1) = 4 (mod 6)
It seems like all the even powers of 2 leave a remainder of 4 when divided by 6.
However, if we subtract one, we would get an odd number for n which goes against the case restrictions. Therefore, there are 0?
Please correct me if I am wrong.
I'll use Sophie Germain's Identity for this problem and don't think it would be posssible to use the Simon's favorite factoring trick. The Simon's favorite factoring trick is only applicable in:
- Two variable problems
- Degree of no more than 2.
z^4 + 4(1^4) = (z^2 + 2z + 2)(z^2 - 2z + 2)
You can find the complex roots by using the quadratic formula from there.
For this problem, you can even use Roots of Unity: https://artofproblemsolving.com/wiki/index.php/Roots_of_unity.