The number $16^4+16^2+1$ is divisible by four distinct prime numbers. Compute the sum of these four primes.

[i]2018 CCA Math Bonanza Lightning Round #3.1[/i]
1. We start with the given expression \(16^4 + 16^2 + 1\). Notice that \(16 = 2^4\), so we can rewrite the expression in terms of powers of 2:
   \[
   16^4 + 16^2 + 1 = (2^4)^4 + (2^4)^2 + 1 = 2^{16} + 2^8 + 1
   \]

2. Next, we recognize that \(2^{16} + 2^8 + 1\) can be factored using the sum of cubes formula. Recall that \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)\) when \(a + b + c = 0\). However, in this case, we use a different approach:
   \[
   2^{16} + 2^8 + 1 = (2^8 + 1)^2 - (2^4)^2
   \]

3. We recognize this as a difference of squares:
   \[
   (2^8 + 1)^2 - (2^4)^2 = (2^8 + 1 + 2^4)(2^8 + 1 - 2^4)
   \]

4. Simplify the factors:
   \[
   2^8 = 256, \quad 2^4 = 16
   \]
   \[
   2^8 + 1 = 257, \quad 2^8 + 1 + 2^4 = 257 + 16 = 273, \quad 2^8 + 1 - 2^4 = 257 - 16 = 241
   \]

5. Thus, we have:
   \[
   2^{16} + 2^8 + 1 = 273 \times 241
   \]

6. Now, we factorize 273 and 241:
   \[
   273 = 3 \times 7 \times 13
   \]
   \[
   241 \text{ is a prime number}
   \]

7. Therefore, the prime factors of \(16^4 + 16^2 + 1\) are \(3, 7, 13,\) and \(241\).

8. The sum of these prime factors is:
   \[
   3 + 7 + 13 + 241 = 264
   \]

The final answer is \(\boxed{264}\).