Positive integers $a$ and $b$ satisfy $a^3 + 32b + 2c = 2018$ and $b^3 + 32a + 2c = 1115$. Find $a^2 + b^2 + c^2$.
1. We start with the given equations:
   \[
   a^3 + 32b + 2c = 2018
   \]
   \[
   b^3 + 32a + 2c = 1115
   \]

2. Subtract the second equation from the first:
   \[
   (a^3 + 32b + 2c) - (b^3 + 32a + 2c) = 2018 - 1115
   \]
   Simplifying, we get:
   \[
   a^3 - b^3 + 32b - 32a = 903
   \]

3. Factor the left-hand side:
   \[
   a^3 - b^3 + 32b - 32a = (a - b)(a^2 + ab + b^2) + 32(b - a)
   \]
   \[
   = (a - b)(a^2 + ab + b^2 - 32)
   \]

4. Set the factored form equal to 903:
   \[
   (a - b)(a^2 + ab + b^2 - 32) = 903
   \]

5. Consider the factors of 903. The prime factorization of 903 is:
   \[
   903 = 3 \times 3 \times 100 + 3 = 3 \times 301
   \]
   The factors of 903 are: \(1, 3, 7, 21, 43, 129, 301, 903\).

6. We test the factor pairs \((a - b, a^2 + ab + b^2 - 32)\). We start with \((3, 301)\):
   \[
   a - b = 3 \quad \text{and} \quad a^2 + ab + b^2 - 32 = 301
   \]

7. Solve for \(a\) and \(b\):
   \[
   a = b + 3
   \]
   Substitute \(a = b + 3\) into \(a^2 + ab + b^2 - 32 = 301\):
   \[
   (b + 3)^2 + (b + 3)b + b^2 - 32 = 301
   \]
   \[
   b^2 + 6b + 9 + b^2 + 3b + b^2 - 32 = 301
   \]
   \[
   3b^2 + 9b - 23 = 301
   \]
   \[
   3b^2 + 9b - 324 = 0
   \]
   Divide by 3:
   \[
   b^2 + 3b - 108 = 0
   \]

8. Solve the quadratic equation:
   \[
   b = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-108)}}{2 \cdot 1}
   \]
   \[
   b = \frac{-3 \pm \sqrt{9 + 432}}{2}
   \]
   \[
   b = \frac{-3 \pm \sqrt{441}}{2}
   \]
   \[
   b = \frac{-3 \pm 21}{2}
   \]
   \[
   b = 9 \quad \text{or} \quad b = -12
   \]
   Since \(b\) is a positive integer, we have:
   \[
   b = 9
   \]

9. Substitute \(b = 9\) back to find \(a\):
   \[
   a = b + 3 = 9 + 3 = 12
   \]

10. Substitute \(a = 12\) and \(b = 9\) back into the original equations to find \(c\):
    \[
    12^3 + 32 \cdot 9 + 2c = 2018
    \]
    \[
    1728 + 288 + 2c = 2018
    \]
    \[
    2016 + 2c = 2018
    \]
    \[
    2c = 2
    \]
    \[
    c = 1
    \]

11. Finally, calculate \(a^2 + b^2 + c^2\):
    \[
    a^2 + b^2 + c^2 = 12^2 + 9^2 + 1^2 = 144 + 81 + 1 = 226
    \]

The final answer is \(\boxed{226}\).