Prove that the equation m^2 − 2013 = 2^n has no solutions in natural numbers.

To prove that the equation \( m^2 - 2013 = 2^n \) has no solutions in natural numbers, we can use properties of modular arithmetic and parity (odd and even numbers).

Consider the equation:

\[ m^2 - 2013 = 2^n \]

Rewriting this, we get:

\[ m^2 = 2^n + 2013 \]

**Step 1: Analyze \( m^2 \mod 4 \)**

- \( m^2 \) can be either \( 0 \) or \( 1 \mod 4 \) because a square number is always congruent to 0 or 1 modulo 4.
- \( 2^n \) modulo 4:
  - If \( n = 0 \), \( 2^n = 1 \mod 4 \)
  - If \( n \geq 2 \), \( 2^n = 0 \mod 4 \)
- 2013 modulo 4:
  - \( 2013 \equiv 1 \mod 4 \)

Now consider the equation \( m^2 = 2^n + 2013 \mod 4 \):

- If \( n = 0 \): \( 2^0 = 1 \Rightarrow 2^0 + 2013 \equiv 1 + 1 = 2 \mod 4 \), and 2 is not a quadratic residue modulo 4. So \( n \neq 0 \).
- If \( n \geq 2 \): \( 2^n \equiv 0 \mod 4 \), \( 2^n + 2013 \equiv 0 + 1 = 1 \mod 4 \)

Therefore, for \( n \geq 2 \):

\[ m^2 \equiv 1 \mod 4 \]

**Step 2: Analyze \( m^2 \mod 8 \)**

Next, we consider higher powers of \( 2 \). Specifically, look at modulo 8 to further restrict \( n \).

- \( m^2 \mod 8 \):
  - \( m^2 \) can be \( 0, 1, 4 \mod 8 \) because a square number modulo 8 results in these values.
- \( 2^n \mod 8 \):
  - If \( n = 1 \): \( 2^1 \equiv 2 \mod 8 \)
  - If \( n \geq 3 \

Explore the proof that the equation m^2 - 2013 = 2^n has no solutions in natural numbers using modular arithmetic and parity analysis. This problem involves understanding quadratic residues and modulo properties, suitable for advanced high school students and beyond.

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