Math Problem Statement

you can use the Euclidean Algorithm to find the gcd

An integer a is even if a = 2k for some integer k. Likewise, a is odd if a = 2k + 1 for some integer k. Prove that gcd(a, a + 2) = 1 if a is odd and gcd(a, a + 2) = 2 if a is even.

Solution

Let's work through the problem step by step using the Euclidean Algorithm.

Problem Breakdown

We are given:

  1. An integer aa is even if a=2ka = 2k for some integer kk.
  2. An integer aa is odd if a=2k+1a = 2k + 1 for some integer kk.

We need to prove:

  • gcd(a,a+2)=1\gcd(a, a+2) = 1 if aa is odd.
  • gcd(a,a+2)=2\gcd(a, a+2) = 2 if aa is even.

Proof

We'll consider both cases for aa: odd and even.

Case 1: aa is Odd

If aa is odd, we can write a=2k+1a = 2k + 1 for some integer kk. Then: a+2=(2k+1)+2=2k+3.a + 2 = (2k + 1) + 2 = 2k + 3. Thus, we have:

  • a=2k+1a = 2k + 1 (an odd number),
  • a+2=2k+3a + 2 = 2k + 3 (also an odd number).

Now we want to find gcd(a,a+2)=gcd(2k+1,2k+3)\gcd(a, a+2) = \gcd(2k+1, 2k+3).

Using the Euclidean Algorithm: gcd(2k+1,2k+3)=gcd(2k+1,(2k+3)(2k+1))=gcd(2k+1,2).\gcd(2k+1, 2k+3) = \gcd(2k+1, (2k+3) - (2k+1)) = \gcd(2k+1, 2). Since 2k+12k+1 is odd, it is not divisible by 2, so the only common divisor is 1. Therefore: gcd(a,a+2)=1.\gcd(a, a+2) = 1. This proves the first part: if aa is odd, gcd(a,a+2)=1\gcd(a, a+2) = 1.

Case 2: aa is Even

If aa is even, we can write a=2ka = 2k for some integer kk. Then: a+2=2k+2=2(k+1).a + 2 = 2k + 2 = 2(k + 1). Thus, we have:

  • a=2ka = 2k (an even number),
  • a+2=2(k+1)a + 2 = 2(k + 1) (also an even number).

Now we want to find gcd(a,a+2)=gcd(2k,2k+2)\gcd(a, a+2) = \gcd(2k, 2k+2).

Using the Euclidean Algorithm: gcd(2k,2k+2)=gcd(2k,(2k+2)2k)=gcd(2k,2).\gcd(2k, 2k+2) = \gcd(2k, (2k+2) - 2k) = \gcd(2k, 2). Since 2k2k is divisible by 2, we can factor out 2 from both terms: gcd(2k,2)=2.\gcd(2k, 2) = 2. Therefore: gcd(a,a+2)=2.\gcd(a, a+2) = 2. This proves the second part: if aa is even, gcd(a,a+2)=2\gcd(a, a+2) = 2.

Conclusion

We have shown that:

  • If aa is odd, gcd(a,a+2)=1\gcd(a, a+2) = 1.
  • If aa is even, gcd(a,a+2)=2\gcd(a, a+2) = 2.

Would you like more details on any part of this solution, or have any questions?


Here are five related questions that might extend your understanding:

  1. What is the Euclidean Algorithm and how is it used in finding the greatest common divisor?
  2. How does the parity (odd or even nature) of numbers affect the gcd calculations?
  3. Can you find the gcd of numbers with a larger difference, such as aa and a+4a+4?
  4. What would the gcd of aa and a+1a+1 be in both odd and even cases?
  5. How does modular arithmetic simplify the proof of gcd properties?

Tip: When proving properties involving gcd, consider using the Euclidean Algorithm, as it simplifies finding common factors between two numbers.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Greatest Common Divisor (GCD)
Parity of Integers
Euclidean Algorithm

Formulas

gcd(a, b) using the Euclidean Algorithm

Theorems

Properties of GCD
Parity of integers

Suitable Grade Level

Grades 11-12