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Prove that there is no $x,y$ rationals such that $x^2 + y^2 = 3$.

Let's assume for the sake of contradiction that this statement is false, i.e. that there exists $x,y \in \mathbb{Q}$ satisfying $x^2 + y^2 = 3$. This is equivalent to: $$\exists(p,q) \in \mathbb{Z}\times \mathbb{N}^*, \exists (p',q') \in \mathbb{Z}\times \mathbb{N}^*: \left(\frac{p}{q}\right)^2 + \left(\frac{p'}{q'}\right)^2 = 3 \wedge \gcd(p,q) = 1 \wedge \gcd(p',q') = 1$$ We then have: $$(pq')^2 + (p'q)^2 = 3(qq')^2$$ We know that a square is either $0$ or $1 \bmod 4$. $(qq')^2$ cannot be $1 \bmod 4$ because that would imply $3(qq')^2 \equiv 3\pmod4$ whereas $(pq')^2 + (p'q)^2 \bmod 3$ cannot be greater than $2$. So $(pq')^2 \equiv 0 \pmod4$, $(p'q)^2 \equiv 0 \pmod 4$ and $(qq')^2 \equiv 0 \pmod 4$. It follows that $p$ or $q'$ is even, $p'$ or $q$ is even and $q$ or $q'$ is even, thus ensuring the following statement to be true: $$(2 \mid p \vee 2 \mid q') \wedge (2 \mid p' \vee 2 \mid q) \wedge (2 \mid q \vee 2 \mid q')$$

The fact that $\frac{p}{q}$ and $\frac{p'}{q'}$ are irreducible prevents $2$ from dividing both $p$ and $q$ or both $p'$ and $q'$. Thus, only $q$ and $q'$ are even numbers while $p$ and $p'$ are odd. Furthermore, by definition: $$ \begin{align} y^2 & = 3 - x^2 \\ & = \frac{3q^2 - p^2}{q^2} \end{align} $$

Since $p$ is odd and $q$ is even, we have $p^2 \equiv 1 \pmod 4$ and $q^2 \equiv 0 \pmod 4$ so $3q^2 - p^2 \equiv 3 \pmod 4$; consequently, $3q^2 - p^2$ is not a perfect square and we have successively: $$\nexists p' \in \mathbb{Z} : p'^2 = 3q^2 - p^2 \\ \forall q' \in \mathbb{N}^*, \nexists p' \in \mathbb{Z} : y^2 = \frac{p'^2}{q'^2} \\ \nexists (p',q') \in \mathbb{Z} \times \mathbb{N}^* : y = \frac{p'}{q'}$$

Thus contradicting our initial assumption. $\blacksquare$

My questions are: how can I improve the writing of this proof? Are there other, more effective ones?

contrapunctus
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  • Yes, there is a "more effective" proof: if $p\ne 0$, then $\frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power. See this duplicate for an easy proof. Since $3=3^1$ and $1$ is not even, we are done. – Dietrich Burde Oct 25 '24 at 13:37

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Your proof looks fine, I don't see what can be improved. A more effective one:

Assume for contradiction that $x^2 + y^2 = 3$ for some $x, y \in \mathbb{Q}$. Eliminating the denominators, we get $a^2 + b^2 = 3c^2$ for some $a, b, c \in \mathbb{Z}$. Extracting the common factor we can assume that $\operatorname{gcd}(a, b, c) = 1$.

Squares give $0$ or $1 \pmod{3}$, so we must have $a \equiv b \equiv 0 \pmod{3}$, and so $c \equiv 0 \pmod{3}$, contradicting the fact $\operatorname{gcd}(a, b, c) = 1$.

Adayah
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  • I think the statement $c \equiv 0 \pmod 3$ could/should be expanded out a bit: That $a,b$ are both multiples of $3$, gives the sum of the squares of those 2 integers a multiple of $9$, which gives $3c^2$ a multiple of $9$, which gives $c^2$ itself a multiple of $3$, which gives $c$ a multiple of $3$. Anyway it took me a minute to see what was really happening here. – Mike Oct 25 '24 at 16:06
  • I do not understand this sentence: “By extracting the common factor, we can assume that $gcd(a,b,c)=1$”. How can we be sure that when we write $\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 3$, we have $a,b,c$ with no common factor, since both fractions have the same denominator? – contrapunctus Oct 25 '24 at 16:10
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    @Mike Yes, that's the argument I had in mind. It is exactly of the same type as in the proof that $\sqrt{3}$ is irrational, so I assumed it wouldn't cause any trouble. – Adayah Oct 25 '24 at 16:21
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    OP: If $a$, $b$, $c$ do have a common factor $r$, then divide both sides of the equation $a^2+b^2=3c^2$ by $r^2$ [the gcd squared] and work with the smaller integers $\frac{a}{r}$, $\frac{b}{r}$, $\frac{c}{r}$; these are integers that do not have a common factor. – Mike Oct 25 '24 at 16:27
  • That's fair I see how it is a standard argument... – Mike Oct 25 '24 at 16:28
  • @Mike Oh, now I understand. I tried to solve the problem with only $3$ integers, $a,b,c$, and I got the same results about modulo $3$, but I couldn't conclude since for me these integers could have a common factor... I feel a bit silly now! :-) Thank you. – contrapunctus Oct 25 '24 at 16:37
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    Anyways OP your question was a valid and good one....indeed, after eliminating denominators $a$, $b$, and $c$ are not necessarily relatively prime, but whenever given an equation $da^k + eb^k = gc^k$, where $d,e,g, k$ are integers and $k$ is a positive integer, we can divide both sides of this equation by $(\gcd(a,b,c))^k$ to get such an equation where $a,b,c$ are relatively prime. – Mike Oct 25 '24 at 19:00