$p+\sqrt{q}=r+\sqrt{s} \qquad p, q, r, s \in \mathbb{Q}$

Question

We must show that, if $\; p+\sqrt{q}=r+\sqrt{s} \;$ and p, q, r, s are rationals then:

either $\; p=r \; $ and $\; q=s$

or $\; q\;$ and $\; s\;$ are both squares of rationals.

I have squared both sides of the given equation and then rearranged to get:

$\sqrt{sq}=\dfrac{p^{2}-2pr+r^{2}-s-q}{2} \; \in \mathbb{Q}$

$\; \sqrt{sq} \; \in \mathbb{Q} \;$ means that $\; sq\;$ is a perfect square but I don't know how to show it must hold for both of them separately.

Answer 1

Suppose that $sq$ is a perfect square and that $s\neq q$. Then $\frac sq$ is also a perfect square (since it's equal to $\frac{sq}{q^2}$). Let $m\in\mathbb{Q}_+$ such that $m^2=\frac sq$. Since $s\neq q$, $m\neq1$. On the other hand$$p+\sqrt q=r+\sqrt s=r+m\sqrt q$$and therefore$$\sqrt q=\frac{p-r}{m-1}.$$But then $q$ is a perfect square and, since $\frac sq$ is a perfect square, $s$ is a perfect square too.

Answer 2

Multiply the equation by $r-\sqrt s$ to obtain $$ pr-p\sqrt s+r\sqrt q-\sqrt{qs}=r^2-s,$$ or rearranged, $$\tag1\sqrt{qs}=pr+s-r^2-p\sqrt s+r\sqrt q .$$ Similarly, multiply the original equation by $p-\sqrt q$ and rearranging (or simply by simmetry $p\leftrightarrow r$, $q\leftrightarrow s$), we obtain $$\tag2\sqrt{qs}=pr+q-p^2-r\sqrt q+p\sqrt s .$$ By equating these expressions for $\sqrt{qs}$, $$ pr+s-r^2-p\sqrt s+r\sqrt q =pr+q-p^2-r\sqrt q+p\sqrt s$$ or, $$\tag3 s-q+p^2-r^2 =-2r\sqrt q+2p\sqrt s$$ Together with the rearranged original equation $$\tag4 p-r=\sqrt s-\sqrt q$$ note that $(3)-2r(4)$ gives $$ s-q+p^2+r^2 -2rp=2(p-r)\sqrt s$$ and $(3)-2p(4)$ gives $$ s-q-p^2-r^2+2pr =2(p-r)\sqrt q$$ If $p\ne r$, these equations show that $\sqrt s$ and $\sqrt q$ are rational.

Answer 3

Let $k=p-r$. Then $k\in\mathbb{Q}$ and $k+\sqrt{q}=\sqrt{s}$.

\begin{align*} k^2+2k\sqrt{q}+q&=s\\ \end{align*}

If $k\ne 0$, then $\displaystyle \sqrt{q}=\frac{s-q-k^2}{2k}\in\mathbb{Q}$ and $\sqrt{s}=k+\sqrt{q}\in\mathbb{Q}$.

If $k=0$, then $p=r$ and $\sqrt{q}=\sqrt{s}$. Hence, $q=s$.