Number Theory ### Educational Resource: Mathematical Proofs#### Problem Statement:- **Objective:** Show that if \( a \) is an odd integer, then \( a^2 \equiv 1 \ (\text{mod} \ 16) \).This problem requires us to prove that for any odd integer \( a \), the square of \( a \) is congruent to 1 modulo 16. This is a very intriguing property of odd integers and congruence relationships that is fundamental in number theory. We can approach this problem using basic properties of odd numbers and modular arithmetic.#### Steps for the Proof:1. **Expression for Odd Integers:** - An odd integer \( a \) can be expressed in the form \( a = 2k + 1 \), where \( k \) is an integer. 2. **Squaring the Odd Integer:** - Square \( a \): \[ a^2 = (2k + 1)^2 = 4k^2 + 4k + 1 \] 3. **Modulo 16 Calculation:** - Simplify \( 4k^2 + 4k + 1 \) modulo 16: \[ a^2 \equiv 4k^2 + 4k + 1 \ (\text{mod} \ 16) \]4. **Modulo Properties:** - Notice that both \( 4k^2 \) and \( 4k \) are multiples of 4. \[ 4k^2 \equiv 0 \ (\text{mod} \ 4) \quad \text{and} \quad 4k \equiv 0 \ (\text{mod} \ 4) \]5. **Summing the Components:** - Thus, the sum \( 4k^2 + 4k \equiv 0 \ (\text{mod} \ 16) \): \[ a^2 \equiv 0 + 0 + 1 \ (\text{mod} \ 16) \quad \Rightarrow \quad a^2 \equiv 1 \ (\text{mod} \ 16) \]#### Conclusion:Therefore, we have shown that for any odd integer \( a \), it holds true that \( a^2 \

Name: Number Theory ### Educational Resource: Mathematical Proofs#### Proble
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: Number Theory ### Educational Resource: Mathematical Proofs#### Problem Statement:- **Objective:** Show that if \( a \) is an odd integer, then \( a^2 \equiv 1 \ (\text{mod} \ 16) \).This problem requires us to prove that for any odd integer \( a \),