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Prove or disprove the statement. For all integers x, y and z, if x|y and x|z, then x|( ) (6𝑦2 − 𝑧2

Question

Prove or disprove the statement.

For all integers x, y, and z, if

  • x | y
  • x | z

then

x | (6y² - z²).

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Solution

To prove the statement, we will follow a structured approach.

1. Break Down the Problem

We need to prove that if x x divides y y and z z (denoted as xy x|y and xz x|z ), then x x must also divide the expression 6y2z2 6y^2 - z^2 .

2. Relevant Concepts

According to the definition of divisibility, if xy x|y , then there exists an integer k k such that: y=kx(1) y = kx \quad \text{(1)} Similarly, if xz x|z , then: z=mx(2) z = mx \quad \text{(2)} for some integer m m .

3. Analysis and Detail

Next, we substitute equations (1) and (2) into the expression 6y2z2 6y^2 - z^2 .

Step 3.1: Substitute y y and z z

Using y=kx y = kx and z=mx z = mx : 6y2z2=6(kx)2(mx)2 6y^2 - z^2 = 6(kx)^2 - (mx)^2

Step 3.2: Expand the terms

Now, we can simplify: 6(kx)2=6k2x2 6(kx)^2 = 6k^2x^2 (mx)2=m2x2 (mx)^2 = m^2x^2 Thus: 6y2z2=6k2x2m2x2 6y^2 - z^2 = 6k^2x^2 - m^2x^2

Step 3.3: Factor the expression

Now, we can factor out x2 x^2 : =(6k2m2)x2 = (6k^2 - m^2)x^2

4. Verify and Summarize

Since 6k2m2 6k^2 - m^2 is an integer (as both k k and m m are integers), we see that: 6y2z2=(6k2m2)x2 6y^2 - z^2 = (6k^2 - m^2)x^2

This means x x divides 6y2z2 6y^2 - z^2 . Therefore, since x x divides (6k2m2)x2 (6k^2 - m^2)x^2 , we can conclude: x(6y2z2) x|(6y^2 - z^2)

Final Answer

The statement is true: For all integers x x , y y , and z z , if xy x|y and xz x|z , then x(6y2z2) x|(6y^2 - z^2) .

This problem has been solved

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