If $a,b,c$ are the roots of $x^3+20x^2+1x+5$, compute $(a^2+1)(b^2+1)(c^2+1)$.

[i]2015 CCA Math Bonanza Tiebreaker Round #2[/i]
1. Given that \(a, b, c\) are the roots of the polynomial \(x^3 + 20x^2 + x + 5\), we need to compute \((a^2 + 1)(b^2 + 1)(c^2 + 1)\).

2. Notice that \((a^2 + 1)(b^2 + 1)(c^2 + 1)\) can be rewritten using complex numbers. Specifically, we can use the fact that \(a^2 + 1 = (a - i)(a + i)\), \(b^2 + 1 = (b - i)(b + i)\), and \(c^2 + 1 = (c - i)(c + i)\).

3. Therefore, \((a^2 + 1)(b^2 + 1)(c^2 + 1) = (a - i)(a + i)(b - i)(b + i)(c - i)(c + i)\).

4. Given the polynomial \(f(x) = x^3 + 20x^2 + x + 5\), we know that \(f(x) = (x - a)(x - b)(x - c)\).

5. To find \((a - i)(a + i)(b - i)(b + i)(c - i)(c + i)\), we evaluate \(f(i)\) and \(f(-i)\):
   \[
   f(i) = i^3 + 20i^2 + i + 5 = -i - 20 - i + 5 = -2i - 15
   \]
   \[
   f(-i) = (-i)^3 + 20(-i)^2 - i + 5 = i - 20 + i + 5 = 2i - 15
   \]

6. The product \((a - i)(a + i)(b - i)(b + i)(c - i)(c + i)\) is equal to \(f(i) \cdot f(-i)\):
   \[
   f(i) \cdot f(-i) = (-2i - 15)(2i - 15)
   \]

7. Simplify the product:
   \[
   (-2i - 15)(2i - 15) = (-2i)(2i) + (-2i)(-15) + (-15)(2i) + (-15)(-15)
   \]
   \[
   = -4i^2 + 30i - 30i + 225
   \]
   \[
   = -4(-1) + 225
   \]
   \[
   = 4 + 225
   \]
   \[
   = 229
   \]

The final answer is \(\boxed{229}\).