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We are given the quadratic congruence:

\[
2x^2 + 2x + 6 \equiv 0 \ (\text{mod} \ 31)
\]

### Step-by-step solution:

1. **Simplify the equation modulo 31**:
   Since 31 is prime, we first divide the entire equation by 2 (since 2 has an inverse modulo 31). To do this, multiply through by the modular inverse of 2 modulo 31. The inverse of 2 modulo 31 can be calculated as follows:
   
   \[
   2^{-1} \ (\text{mod} \ 31)
   \]
   Using the extended Euclidean algorithm, we find that \( 2^{-1} \equiv 16 \ (\text{mod} \ 31) \).

2. **Apply the inverse**:
   Multiply the equation \( 2x^2 + 2x + 6 \equiv 0 \ (\text{mod} \ 31) \) by 16:
   
   \[
   16 \cdot (2x^2 + 2x + 6) \equiv 0 \ (\text{mod} \ 31)
   \]
   This simplifies to:
   
   \[
   32x^2 + 32x + 96 \equiv 0 \ (\text{mod} \ 31)
   \]

3. **Reduce modulo 31**:
   Now reduce each coefficient modulo 31:
   \[
   32x^2 \equiv x^2 \ (\text{mod} \ 31)
   \]
   \[
   32x \equiv x \ (\text{mod} \ 31)
   \]
   \[
   96 \equiv 4 \ (\text{mod} \ 31)
   \]
   Therefore, the equation becomes:
   
   \[
   x^2 + x + 4 \equiv 0 \ (\text{mod} \ 31)
   \]

4. **Solve the quadratic equation**:
   Now we solve \( x^2 + x + 4 \equiv 0 \ (\text{mod} \ 31) \). This is a standard quadratic congruence of the form \( ax^2 + bx + c = 0 \). We can solve it using the quadratic formula in modulo arithmetic:
   
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \ (\text{mod} \ 31)
   \]
   For the equation \( x^2 + x + 4 = 0 \), we have:
   \[
   a = 1, \ b = 1, \ c = 4
   \]
   The discriminant is:
   \[
   \Delta = b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = -15 \ (\text{mod} \ 31)
   \]
   Now reduce \(-15 \mod 31\):
   \[
   -15 \equiv 16 \ (\text{mod} \ 31)
   \]
   So the discriminant is \( 16 \).

5. **Find the square root of the discriminant**:
   Next, we need to find the square root of 16 modulo 31. Since 31 is prime, we can test small integers to see which squares give 16:
   \[
   4^2 = 16 \ (\text{mod} \ 31)
   \]
   So \( \sqrt{16} \equiv 4 \ (\text{mod} \ 31) \).

6. **Compute the solutions**:
   Now apply the quadratic formula:
   \[
   x = \frac{-1 \pm 4}{2} \ (\text{mod} \ 31)
   \]
   This gives two possible solutions:
   \[
   x_1 = \frac{-1 + 4}{2} = \frac{3}{2} \ (\text{mod} \ 31)
   \]
   and
   \[
   x_2 = \frac{-1 - 4}{2} = \frac{-5}{2} \ (\text{mod} \ 31)
   \]
   To handle these, we multiply by the inverse of 2 modulo 31, which is 16:
   \[
   x_1 = 3 \cdot 16 = 48 \equiv 17 \ (\text{mod} \ 31)
   \]
   \[
   x_2 = -5 \cdot 16 = -80 \equiv -80 + 3 \cdot 31 = -80 + 93 = 13 \ (\text{mod} \ 31)
   \]

Thus, the two solutions are:
\[
x_1 = 17 \ (\text{mod} \ 31), \quad x_2 = 13 \ (\text{mod} \ 31)
\]

The smallest positive integer solution is **13**.

Would you like more details, or do you have any other questions?

Here are 5 related questions for further exploration:

1. How do you find the modular inverse using the extended Euclidean algorithm?
2. Why is it important that 31 is prime in this problem?
3. Can you generalize the quadratic formula for solving congruences modulo prime numbers?
4. How do you handle negative numbers in modular arithmetic?
5. What are other methods for solving quadratic congruences besides the quadratic formula?

**Tip:** When solving modular equations, especially quadratics, always check for primes, as it allows easier computations like finding inverses and square roots modulo p!

The quadratic congruence 2x^2 + 2x + 6 ≡ 0 (mod 31). Find the smallest positive integer solution.

This problem shows how to solve the quadratic congruence 2x^2 + 2x + 6 ≡ 0 mod 31 using modular arithmetic. Learn key steps like finding modular inverses and using the quadratic formula.

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