If positive integer x satisfies x^2 -4x + 56 ≡ 14 (mod 17). Find minimum value of x.

We need to solve the congruence equation:

\[ x^2 - 4x + 56 \equiv 14 \pmod{17} \]

### Step 1: Simplifying the congruence modulo 17.

First, reduce the coefficients modulo 17:

\[ x^2 - 4x + 56 \equiv 14 \pmod{17} \]

Since \( 56 \div 17 = 3 \) remainder \( 5 \), we have:

\[ 56 \equiv 5 \pmod{17} \]

So the equation becomes:

\[ x^2 - 4x + 5 \equiv 14 \pmod{17} \]

### Step 2: Rearranging the equation.

Subtract 14 from both sides:

\[ x^2 - 4x + 5 - 14 \equiv 0 \pmod{17} \]

This simplifies to:

\[ x^2 - 4x - 9 \equiv 0 \pmod{17} \]

### Step 3: Solving the quadratic congruence.

We need to solve:

\[ x^2 - 4x \equiv 9 \pmod{17} \]

This is equivalent to solving:

\[ x^2 - 4x - 9 \equiv 0 \pmod{17} \]

### Step 4: Using the quadratic formula modulo 17.

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \( x^2 - 4x - 9 \equiv 0 \), we have:

- \( a = 1 \)
- \( b = -4 \)
- \( c = -9 \)

Substituting these values into the formula:

\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-9)}}{2 \times 1} \]

Simplifying inside the square root:

\[ x = \frac{4 \pm \sqrt{16 + 36}}{2} \]

\[ x = \frac{4 \pm \sqrt{52}}{2} \]

### Step 5: Reducing modulo 17.

First, compute \( \sqrt{52} \) modulo 17. Since \( 52 \div 17 = 3 \) remainder \( 1 \), we have:

\[ 52 \equiv 1 \pmod{17} \]

So, we need to compute \( \sqrt{1} \pmod{17} \), which is either \( 1 \) or \( 16 \) modulo 17.

### Step 6: Finding the solutions.

Substituting these values back into the quadratic formula:

1. For \( \sqrt{1} \equiv 1 \), we get:

   \[ x = \frac{4 \pm 1}{2} \]

   This gives solutions \( x = \frac{5}{2} \equiv 9 \pmod{17} \) and \( x = \frac{3}{2} \equiv 2 \pmod{17} \).

2. For \( \sqrt{1} \equiv 16 \), we get:

   \[ x = \frac{4 \pm 16}{2} \]

   This gives solutions \( x = \frac{20}{2} \equiv 3 \pmod{17} \) and \( x = \frac{-12}{2} \equiv 11 \pmod{17} \).

### Step 7: Choosing the minimum value.

The possible solutions are \( x \equiv 2, 9, 3, 11 \pmod{17} \). The minimum value is \( x = 2 \).

### Final Answer

The minimum value of \( x \) is \( \boxed{2} \).

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Would you like more details or have any questions?

Here are 5 related questions:
1. How do you solve quadratic congruences using the Chinese Remainder Theorem?
2. What is the significance of modulo arithmetic in cryptography?
3. Can all quadratic equations be solved modulo any integer?
4. How can you generalize the quadratic formula for different moduli?
5. What is Hensel’s lemma and how is it applied in number theory?

**Tip:** Always reduce large coefficients modulo the given modulus early to simplify your calculations.

Learn how to solve the quadratic congruence x^2 - 4x + 56 ≡ 14 (mod 17) step-by-step. This problem involves modular arithmetic and the application of the quadratic formula modulo 17. Suitable for advanced high school students and learners interested in number theory.

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