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This question already has an answer here:

Are there rational numbers such that $x^2 + y^2 = 3$ ?


If I want to find a rational paramterizatio of $x^2 + y^2 = 1$ could start with the point $(1,0)$ and find lines $\ell$ of slope $m \in \mathbb{Q}$ and the intersection points $[\ell] \cdot [circle] = 2 [pt] $.

However, if I use the circle $x^2 + y^2 = 2$ there's no rational point on the axes. Instead we should use $(x,y) = (1,1)$.

In the case of $x^2 + y^2 = 3$ there's no obvious rational point that comes to mind. I'm concerned there might be no rational point at all. In integers we'd have $a^2 + b^2 = 3c^2$ with $a,b,c \in \mathbb{Z}$. We'd have $c \equiv 0 \pmod 4$. Then $a \equiv b \equiv 0 \pmod 4$. This could lead to an infinite descent argument.


As a bonus could there exist a small rationqal $\epsilon > 0$ with $\epsilon \ll 1$ and $\epsilon \in \mathbb{Q}$ such that $x^2 + y^2 = 3 + \epsilon$ has a solution (and therefore infinitely many solutions)?

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marked as duplicate by Andres Mejia, Community Jul 9 '18 at 17:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You can assume that the curve has a Rational Point let $$P(\frac{a}{b},\frac{c}{d})$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 9 '18 at 17:29
  • $\begingroup$ $1249^2 + 1200^2 = 3000001 \; . \; \;$ $ 1.249^2 + 1.2^2 = 3.000001 \; \; .$ $\endgroup$ – Will Jagy Jul 9 '18 at 18:28
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Hint: Show that if $a^2+b^2=3c^2$ for some integers $a,b,c$ with $c\neq 0.,$ then it has a solution with neither $a$ nor $b$ divisible by $3$.

Then show that means there must be an integer $n$ such that $n^2+1$ is divisible by $3$.

Is that possible?

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Any integer square is of form $3k$ or $3k+1$ for some integer $k$. Then $3|a^2+b^2$ if both $a$ and $b$ are divisible by $3$ which turns out $a=3a_1$ and $b=3b_1$ and by substitution $$9a_1^2+9b_1^2=3c^2\\3a_1^2+3b_1^2=c^2$$which leads to $c=3c_1$ and we have $$a_1^2+b_1^2=3c_1^2$$ and this procedure continues infinitely many times, then such integers don't exist

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Your concerns are well justified. We will prove that there cannot be any non-trivial (i.e. a non zero) integer solution to the diophantine equation $ x^2 +y^2 = pz^2$ where $ p \equiv 3$ (mod $4$). As a corollary, there cannot be any rational points that satisfy the equation $ x^2 +y^2 = 3z^2$ (why?).

Let us assume we CAN find a non-trivial solution to $ x^2 +y^2 = pz^2$. We will derive a contradiction. Without loss of generality, we may assume that $x,y $ and $z$ have no common divisors i.e. gcd$(x,y,z) = 1$. Otherwise, we may always divide through by the common divisor to obtain another non-trivial solution.

Next, note that a square of an integer is either congruent to $0$ or $1$ (mod$4$). This may be seen by squaring the expressions $2k$ and $2k+1$ respectively.

Hence,

$ x^2 \equiv 0,1 $ (mod$4$),

$ y^2 \equiv 0,1 $ (mod$4$),

$ pz^2 \equiv 0,3 $ (mod$4$).

(listing all posible options).

Thus,

$x^2 +y^2 \equiv 0,1,2 $ (mod$4$).

Therefore, as equality implies equality in congruence, it follows the only possible way for $x^2 +y^2 \equiv pz^2$ is for $x, y$ and $z$ to be even.

This contradicts the fact that we assumed gcd$(x,y,z) = 1$.

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