Algebra proof within relation problem

Question

I need to show an equivalence relation on $ℕ$ defined as follows:

$x$ ~ $y$ ⇔ $x \times y$ is a square

Reflexivity: $x \times x$ is a square, so $x$ ~ $x$, so ~ is reflexive

Symmetry: $x$ ~ $y$ implies $y$ ~ $x$, $x\times y$ is a square, so is $y \times x$

But for transitivity I need to show that if $xy$ and $yz$ are both squares, then $xz$ is also a square.

So I did this:

$xy = a^2$ and $yz = b^2$, then $xz = \frac{a^2\times b^2}{y^2}$ and $xz = (\frac{ab}{y})^2$

But I don't think this works because there is a fraction. How else can I show that $xz$ is a square?

Answer 1

$$ xz y^2 = a^2 b^2 $$

So we know $a^2 b^2$ is divisible by $y^2$. In other words, $\dfrac{ab}{y}$ is an integer. So your solution is correct.

Answer 2

Let $x = \prod_{i=1}^n p_i^{c_i}$, $y = \prod_{j=1}^n p_j^{d_j}$ and $z = \prod_{k=1}^n p_k^{f_k}$, where $P = \{p_i \mid 1\leq i \leq n\}$ is a set of primes containing the primes that divide any of $x$, $y$, or $z$. Then, since $xy$ and $yz$ are squares, writing the factorizations of $xy$ and $yz$ into primes in $P$, we obtain the relations \begin{align*} c_i + d_i &\cong 0 \pmod{2} \\ d_i + f_i &\cong 0 \pmod{2} \end{align*} for each $i \in [1,n]$. Then \begin{align*} (c_i + d_i) + (d_i + f_i) &\cong 0 \pmod{2} \\ c_i + 2d_i + f_i &\cong 0 \pmod{2} \\ c_i + f_i &\cong 0 \pmod{2} \text{,} \end{align*} which says $xz$ is a square.