What about this one?

Consider the problem modulo 3. Since $4 \equiv 1$ and $545 \equiv 2$ mod 3, we have $4^{545} + {545}^5 \equiv 1^{545} + 2^5 \equiv 1 + 2 \equiv 0\ \text{mod 3}$ so that this sum is divisible by 3. Since it is also greater than 3, it is therefore not prime.

Since

$x^5+y^5 = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)$

We just substitute $x = 4^{109}$ and $y = 545$ , hence it is proved that the number above is NOT prime.

Of course, the answer is not yes, so it must be no.