Show that the curve $x^2+y^2-3=0$ has no rational points

Question

Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.

Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.

Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.

Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.

In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.

Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.

Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.

Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.

We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.

Answer 1

Suppose to the contrary that there is a rational solution of the equation. Then there exist integers $a$, $b$ and $q$, with $q\ne 0$, such that $a^2+b^2=3q^2$, and $a$, $b$ and $q$ have no common factor greater than $1$.

Note that $a$ and $b$ must both be divisible by $3$. For if an integer $m$ is not divisible by $3$, then $m^2$ has remainder $1$ on division by $3$. So if one or both of $a$ and $b$ is not divisible by $3$, then $a^2+b^2$ has remainder $1$ or $2$ on division by $3$, and therefore cannot be of the shape $3q^2$.

Thus both $a$ and $b$ are divisible by $3$. It follows that $q$ is divisible by $3$, contradicting our assumption that $a$, $b$ and $q$ have no common divisor greater than $1$.

Answer 2

Suppose $a^2 + b^2 = 3 c^2$ Write $a = 3^p u$, $b = 3^q v$, and $c = 3^r w$, where $u, v, w$ are all relatively prime to 3. This does not assume that $a, b, c$ have no common divisor greater than 1.

Assume $p \le q$ (if $p > q$, switch their roles in what follows). $a^2+b^2 = (3^p u)^2 + (3^q v)^2 = 3^{2p}(u^2+ 3^{2(q-p)}v^2) $, so an even power of 3 divides $a^2+b^2$. ($u^2+ 3^{2(q-p)}v^2$ has a remainder of 1 or 2 mod 3 depending on if $p < q$ or $p = q$.)

But $3 c^2 = 3 (3^r w)^2 = 3^{2r+1} w^2$, so an odd power of 3 divides $c^2$.

By unique prime factorization, this is a contradiction.

Note: I wrote this because the assumption of $a, b, c$ having no common factor is, to me, either an implicit use of unique factorization or an infinite descent contradiction based on the powers of 3 dividing them.

Answer 3

Kind of late but here is a classical proof by contradiction:

Suppose the provided equation has a rational point

$x, y \in \mathbb{Q} \implies x = \frac{a}{b} \text{ and } y = \frac{c}{d} \text{ where gcd(a,b)=1 and gcd(c,d)=1}$

Then the equation becomes: $$ \frac{a^2}{b^2} + \frac{c^2}{d^2} = 3 $$

$$ a^2 = k, \; b^2 = p, \; c^2 = s, \; d^2 = t $$

$$ \frac{k}{p} + \frac{s}{t} = 3 \implies kt + sp = 3pt $$

Now we will have a note: $\forall t \in \mathbb{Z}$ $t^2 \equiv 0 \text{ or } 1 (mod 4)$

Now let's separate into cases:

1. t and p is even
2. all are odd

There can be other cases bu you will see that they are trivial because they either contradict with the gcd=1 rule or modulo rule immediately.

Case 1) This leads to the substitution $ t=4m \text{ and } p=4n \text{ where m and n are also perfect squares }$

$$ 4km + 4sn = 48mn \implies km + sn = 12mn $$ After that it is clear that you can do the exact same substitution for m and n. As you can continue the substitutions all over that leads to a contradiction that t and p, therefore b and d has an infinite amount of 2 multiplied in them, which contradicts the fact that they are integers.

Case 2) Just by looking the equation modulo 2 you will see that there can't be any solution.

All the cases end up with a contradiction which means that our assumption was wrong. Therefore, the equation $x^2+y^2−3=0$ has no rational roots/solutions for both x,y. ■

Answer 4

Let $$a^2+b^2=3q^2\tag{1}$$ we may assume that $3\not\mid a,b$, otherwise $3\mid q$ and so we have an infinite descent. Reducing eq. $(1)$ we have $$a^2+b^2\equiv_3 0$$ This is only possible when $3\mid,a,b$. Contradiction.

Answer 5

Another proof using complex numbers (essentially the same):

Let $a^2+b^2=3q^2$, and suppose $a,b,q$ have no common factor. Consider $$(a+bi)(a-bi)=3q^2\in\Bbb Z[i]\tag{1}$$

$3$ is still a prime in $\Bbb Z[i]$

Proof: Let $3=\alpha\beta$ with $\alpha,\beta\in\Bbb Z[i]$. Then taking norms $$(c^2+d^2)(e^2+f^2)=9\tag{2}$$ Assuming none of $\alpha,\beta$ are units we have $$c^2+d^2=3\\c^2+d^2\equiv_30$$ This is only possible when $3\mid c,d$, likewise $3\mid e,f$. Therefore, from eq. $(2)$, $81\mid 9$. Contradiction.

Since $3\in\Bbb Z[i]$ is prime we have that $3\mid a\pm bi$, but then $3\mid a,b$. This yields an infinite descent. Contradiction.