Math Problem Statement

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

Solution

We need to solve the congruence equation:

x24x+5614(mod17)x^2 - 4x + 56 \equiv 14 \pmod{17}

Step 1: Simplifying the congruence modulo 17.

First, reduce the coefficients modulo 17:

x24x+5614(mod17)x^2 - 4x + 56 \equiv 14 \pmod{17}

Since 56÷17=356 \div 17 = 3 remainder 55, we have:

565(mod17)56 \equiv 5 \pmod{17}

So the equation becomes:

x24x+514(mod17)x^2 - 4x + 5 \equiv 14 \pmod{17}

Step 2: Rearranging the equation.

Subtract 14 from both sides:

x24x+5140(mod17)x^2 - 4x + 5 - 14 \equiv 0 \pmod{17}

This simplifies to:

x24x90(mod17)x^2 - 4x - 9 \equiv 0 \pmod{17}

Step 3: Solving the quadratic congruence.

We need to solve:

x24x9(mod17)x^2 - 4x \equiv 9 \pmod{17}

This is equivalent to solving:

x24x90(mod17)x^2 - 4x - 9 \equiv 0 \pmod{17}

Step 4: Using the quadratic formula modulo 17.

The quadratic formula is given by:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation x24x90x^2 - 4x - 9 \equiv 0, we have:

  • a=1a = 1
  • b=4b = -4
  • c=9c = -9

Substituting these values into the formula:

x=4±(4)24×1×(9)2×1x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-9)}}{2 \times 1}

Simplifying inside the square root:

x=4±16+362x = \frac{4 \pm \sqrt{16 + 36}}{2}

x=4±522x = \frac{4 \pm \sqrt{52}}{2}

Step 5: Reducing modulo 17.

First, compute 52\sqrt{52} modulo 17. Since 52÷17=352 \div 17 = 3 remainder 11, we have:

521(mod17)52 \equiv 1 \pmod{17}

So, we need to compute 1(mod17)\sqrt{1} \pmod{17}, which is either 11 or 1616 modulo 17.

Step 6: Finding the solutions.

Substituting these values back into the quadratic formula:

  1. For 11\sqrt{1} \equiv 1, we get:

    x=4±12x = \frac{4 \pm 1}{2}

    This gives solutions x=529(mod17)x = \frac{5}{2} \equiv 9 \pmod{17} and x=322(mod17)x = \frac{3}{2} \equiv 2 \pmod{17}.

  2. For 116\sqrt{1} \equiv 16, we get:

    x=4±162x = \frac{4 \pm 16}{2}

    This gives solutions x=2023(mod17)x = \frac{20}{2} \equiv 3 \pmod{17} and x=12211(mod17)x = \frac{-12}{2} \equiv 11 \pmod{17}.

Step 7: Choosing the minimum value.

The possible solutions are x2,9,3,11(mod17)x \equiv 2, 9, 3, 11 \pmod{17}. The minimum value is x=2x = 2.

Final Answer

The minimum value of xx is 2\boxed{2}.


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.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Quadratic Congruences

Formulas

Quadratic formula modulo n

Theorems

Chinese Remainder Theorem

Suitable Grade Level

Advanced High School