Let's break down the problem and solve it step-by-step **using the contrapositive**. ## Original Statement > If \( n^2 - 4n + 3 \) is even, then \( n \) is odd. ## Step 1: Write the Contrapositive The contrapositive of "If P then Q" is "If not Q then not P". So, the contrapositive is: > If \( n \) is **even**, then \( n^2 - 4n + 3 \) is **odd**. If we prove this, we have proved the original statement. --- ## Step 2: Assume \( n \) is Even Let \( n = 2k \) where \( k \) is an integer. --- ## Step 3: Substitute \( n \) into the Expression \[ n^2 - 4n + 3 = (2k)^2 - 4(2k) + 3 \] \[ = 4k^2 - 8k + 3 \] --- ## Step 4: Analyze the Parity (Even or Odd) - \( 4k^2 \) is always **even** (since 4 times any integer squared is even) - \( 8k \) is always **even** - \( 3 \) is **odd** Now, **even - even = even**, and **even + odd = odd**. So: \[ 4k^2 - 8k \text{ is even} \] \[ (\text{even}) + 3 = \text{odd} \] Therefore, \( n^2 - 4n + 3 \) is **odd** when \( n \) is even. --- ## Step 5: Conclusion We have proved the contrapositive: > If \( n \) is even, then \( n^2 - 4n + 3 \) is odd. Therefore, the original statement is **true**: > **If \( n^2 - 4n + 3 \) is even, then \( n \) is odd.** --- ## **Final Answer** **By proving the contrapositive, we have shown that if \( n^2 - 4n + 3 \) is even, then \( n \) must be odd.**