Show that no element $(x,y)\in\mathbb{Q}^{2}$ satisfies $x^{2}+y^{2}-3=0$.

Question

I am going through the book of proof by Richard Hammack, and came across this exercise:

We say that a point $P=(x,y)$ in $\mathbb{R}^{2}$ is rational if both $x$ and $y$ are rational. More precisely, $P$ is rational if $P=(x,y)\in\mathbb{Q}^{2}$. An equation $F(x,y)=0$ is said to have a rational point if there exists $x_0,y_0\in\mathbb{Q}$ such that $F(x_0,y_0)=0$. For example, $x^{2}+y^{2}-1=0$ has a rational point $(x_0,y_0)=(1,0)$. Show that the curve $x^{2}+y^{2}-3=0$ has no rational points.

My proof is the following:

For the sake of contradiction, suppose it is not the case that the curve $x^{2}+y^{2}-3=0$ has no rational points. Thus, let $(x_0,y_0)$ be a rational point of the curve $x^{2}+y^{2}-3=0$.

Hence, $(x_0,y_0)=(\frac{p_1}{q_1},\frac{p_2}{q_2})$, for some $p_1,p_2,q_1,q_2\in\mathbb{Z}$, $q_1,q_2\neq0$, $\textrm{gcd}(p_1,q_1)=1$ and $\textrm{gcd}(p_2,q_2)=1$.

Notice that \begin{align} x^{2}+y^{2}-3&=0\nonumber\\ x^{2}+y^{2}&=3\nonumber\\ \left(\frac{p_1^{2}}{q_1^{2}}\right)+\left(\frac{p_2^{2}}{q_2^{2}}\right)&=3\nonumber\\ p_1^{2}q_2^{2}+p_2^{2}q_1^{2}&=3q_1^{2}q_2^{2}\nonumber \end{align}

Since $q_1^{2}$ divides the right-hand side and the second term in the left-hand side, then it must divide also the first term in the left-hand side. Hence, $q_1^{2}\mid p_1^{2}q_2^{2}$. By hypothesis, $\textrm{gcd}(p_1,q_1)=1$, so $q_1\nmid p_1$. Thus, $q_1^{2}\mid q_2^{2}$.

Similarly, $q_2\mid q_1$. Thus, we conclude $q_1=q_2$. Therefore, \begin{align} p_1^{2}q_2^{2}+p_2^{2}q_1^{2}&=3q_1^{2}q_2^{2}\nonumber\\ p_1^{2}q_1^{2}+p_2^{2}q_1^{2}&=3q_1^{2}q_1^{2}\nonumber\\ p_1^{2}+p_2^{2}&=3q_1^{2}\nonumber \end{align}

So far, so good. I think it is all correct. The following paragraph is where I made some claims with no proof, and dont know if it is ok to do it:

It is known that the quadratics residue modulo $4$ is either $0$ or $1$. Thus, $p_1^{2}+p_2^{2}\equiv 0(\textrm{mod }4)$ or $p_1^{2}+p_2^{2}\equiv 1(\textrm{mod }4)$ or $p_1^{2}+p_2^{2}\equiv 2(\textrm{mod }4)$. But $3q_1^{2}\equiv 0(\textrm{mod }4)$ or $3q_1^{2}\equiv 1(\textrm{mod }4)$.

So, the only possible scenario is the one for which $p_1^{2},p_2^{2},q_1^{2}\equiv 0(\textrm{mod }4)$. In particular, $p_1^{2}\equiv q_1^{2}(\textrm{mod }4)$. Thus, $4$ divides both $p_1^{2}$ and $q_1^{2}$, which means they are both even, which contradicts the fact that $\textrm{gcd}(p_1,q_1)=1$.$ \blacksquare$

So, can I claim that "It is known that the quadratics residue modulo $4$ is either $0$ or $1$" without proof?

Is the claim that "the only possible scenario is the one for which $p_1^{2},p_2^{2},q_1^{2}\equiv 0(\textrm{mod }4)$" obvious enough or should I prove it too?

Answer 1

Use a common denominator to make the notation less ugly: write $x = a/c$ and $y = b/c$ where $c > 0$. We can assume the common denominator is chosen to be as small as possible and will get a contradiction.

The condition $x^2 + y^2 = 3$ in $\mathbf Q$ can be rewritten as $a^2 + b^2 = 3c^2$ in $\mathbf Z$. Now reduce modulo $3$ rather than modulo $4$ as you did: $a^2+ b^2 \equiv 0 \bmod 3$. Let's derive $a \equiv 0 \bmod 3$ and $b \equiv 0 \bmod 3$. If $a \not\equiv 0 \bmod 3$ then from $b^2 \equiv -a^2 \bmod 3$ we get $(b/a)^2 \equiv -1 \equiv 2 \bmod 3$, which is a contradiction since the only squares mod $3$ are $0$ and $1$. Thus $a \equiv 0 \bmod 3$. By swapping the roles of $a$ and $b$, $b \equiv 0 \bmod 3$ too.

Since $a$ and $b$ are multiples of $3$, $a^2 + b^2$ is a multiple of $9$, so $3c^2$ is divisible by $9$, which makes $c$ divisible by $3$. Thus $a$, $b$, and $c$ are all multiples of $3$, which means the fractions $a/c$ and $b/c$ can both be reduced further, so $c$ was not the smallest common denominator. Contradiction!

The exact same reasoning shows for each prime $p \equiv 3 \bmod 4$ that the equation $x^2 + y^2 = p$ has no rational solution because $-1 \bmod p$ is not a square.