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

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

1. Break Down the Problem

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

2. Relevant Concepts

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

3. Analysis and Detail

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

Step 3.1: Substitute $y$ and $z$

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

Step 3.2: Expand the terms

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

Step 3.3: Factor the expression

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

4. Verify and Summarize

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

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

Final Answer

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