Find rational solutions to $x^2 + y^2 = 6$

Question

This question comes from Rational Points on Elliptic Curves (Silverman & Tate) Exercise $1.7$ (a).

Find rational solutions (if any) to $x^2 + y^2 = 6$

I think there exist no solutions and here is my proof:

Suppose that there is a rational point and write it as $$x=\frac{X}{Z} \quad \text { and } \quad y=\frac{Y}{Z}$$ for some integers $X, Y$, and $Z$. Then $$X^2 + Y^2 = 6Z^2$$ If $X, Y, Z$ have a common factor, then we may remove it, so we may assume that they have no common factor. It follows that neither $X$ nor $Y$ is divisible by $6$. This is true because if $6$ were to divide $X$, then $6$ divides $Y^2 = 6Z^2 - X^2$, so $6$ divides $Y$. But then $36$ divides $X^2 + Y^2 = 6Z^2$, so $6$ divides $Z$, contradicting the fact that $X, Y, Z$ have no common factors. Hence $6$ does not divide $X$, and a similar argument shows that $6$ does not divide $Y$.

Since $X$ and $Y$ are not divisible by $6$, we have $$X \equiv \pm 1,4 \quad(\bmod 6) \text { and } Y \equiv \pm 1,4 \quad(\bmod 6),$$ and hence $$X^2 + Y^2 \equiv 1+1 \equiv 2 \quad(\bmod 6)$$ or $$X^2 + Y^2 \equiv 1+4 \equiv 5 \quad(\bmod 6)$$ or $$X^2 + Y^2 \equiv 4+4 \equiv 3 \quad(\bmod 6)$$ However, we also have $$X^2 + Y^2 = 3Z^2 \equiv 0 \quad(\bmod 6),$$ a contradiction.

I think this proof is sufficient. I based it on the method that the book showed there didn't exist any rational points on the curve $x^2 + y^2 = 3$.

Answer 1

The ideas in the proof are correct, but there are some minor errors that need correction to make your reasoning rigorous.

First, while it is true that $X,Z$ and $Y,Z$ may be taken to be relatively prime without loss of generality, and that this directly follows from $x = X/Z$ and $y = Y/Z$, it is not immediately obvious why $X,Y$ need to also be relatively prime. To see why, consider the alternative equation $$X^2 + Y^2 = 40Z^2,$$ which of course admits integer solutions such as $(X,Y,Z) = (2,6,1)$. Then $X, Z$ and $Y, Z$ are relatively prime but $X, Y$ are not. So it is misleading to start off with such a claim; instead, it is better to omit it and directly proceed to your reasoning that neither $X$ nor $Y$ can be divisible by $6$.

Second, you have a few typographical errors:

$$X^2 + Y^2 \equiv 4 + 4 = 8 \equiv \color{red}{2} \pmod 6$$

and your last claim should read

However, we also have $$X^2 + Y^2 = \color{red}{6}Z^2 \equiv 0 \pmod 6.$$

Answer 2

First prove that $x^2 +y^2 =3$ has no solution for $(x,y)\in \mathbb Q^2$. Now, we can see $(x+y)^2 +(x-y) ^2 =6$, put $s=x+y$, $t=x-y$. As $x^2+y^2 =3$ has no solution so $s^2+t^2 =6$ also not has solutions.

Now how to prove $x^2 +y^2 =3$ has no solution? write $x=\frac{p_{1}}{q_{1}}$, $y=\frac{p_{2}}{q_{2}}$. So we get $(p_{1}q_{2})^2 +(q_{1}p_{2})^2 =3q_{1}^2q_{2}^2$ ,this implies 3 divides the lhs, as $n^2$ leaves remainder 0 or 1 for any $n\in \mathbb N$. So,both $(p_{1}q_{2})^2$ and $(q_{1}p_{2})^2$ is divisible by 3, so they are divisible by 9, so lhs divisible by 9, so whenever rhs divisible by $3^{2k-1}$, lhs become divisible by $3^{2k}$, so this implies that no solution exists for $x^2 +y^2 =3$.