Show that if a ú an odd integer, thn a =1 (md 16).
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.
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 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc29e3d0-ab06-4b11-9f3c-9c089e985983%2F8ba7c44a-ae57-4cf6-8ec0-dba56805d921%2F5ibpp17e_processed.jpeg&w=3840&q=75)
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