The statement you've written is:

$\forall x \in \mathbb{Z}, \exists y \in \mathbb{Z} \text{ such that } x = y^2.$

This reads as: "For every $x$ in the set of integers ($\mathbb{Z}$), there exists a $y$ in the set of integers such that $x = y^2$."

This statement is false. Here's why:

Explanation:

1. $y^2$ represents the square of an integer $y$, and squares of integers are called perfect squares (e.g., $0, 1, 4, 9, 16, \dots$).
2. However, not every integer $x$ is a perfect square. For instance:
   • $x = 2$ is not a perfect square because no integer $y$ satisfies $y^2 = 2$.
   • Similarly, integers like $-1$, $-3$, and $-5$ are also not perfect squares.

Thus, the statement implies that every integer $x$ can be written as a square of some integer $y$, but this is not true. Only certain integers (perfect squares) can be written as $y^2$.

Would you like more details or further clarification?

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Here are some related questions to explore further:

1. What integers can be written as squares of other integers?
2. How can we determine if a number is a perfect square?
3. What are the properties of perfect squares within the set of integers?
4. How can we prove that certain numbers are not perfect squares?
5. Are negative numbers ever perfect squares in the set of real numbers?

Tip: Remember that a perfect square is always non-negative, and it comes from multiplying an integer by itself.