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what is the solution of this diophantine equation $x^2+y^2=3$ ? with x and y are the rationnals number I have one solution : $x=\frac{a}{c}$ and $y=\frac{b}{c}$ a, b and c are the positive integer

I found a solution by doing a reasoning modulo 3

Bill Dubuque
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curieux_2014
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  • Aren't only integer solutions allowed for diophantine equations? $\Rightarrow x,y \in \mathbb{Z}$ – mrnovice Feb 27 '17 at 01:52
  • We seek solutions in the set of rational numbers – curieux_2014 Feb 27 '17 at 01:55
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    @mrnovice the homogenous form of this equation is the diophantine equation. $a^2+b^2=3c^2$. – Ali Caglayan Feb 27 '17 at 01:57
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    The comment of @AliCaglayan contains the key to the argument that there are no rational solutions. Indeed, if $x=m/n$ and $y=p/q$ give a solution, then we can multiply both sides of the equation by $n^2q^2$ to get $m^2q^2+p^2n^2=3n^2q^2$. But this contradicts the known fact that a sum of two integer squares has evenly many factors of any prime $\pi\equiv3\pmod4$. – Lubin Feb 27 '17 at 02:43
  • merci cela répond à ma question – curieux_2014 Sep 04 '23 at 18:29

2 Answers2

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Diofantine equation $$x^2 + y^2 = 3z^2$$ has only the trivial (zero) integer solution, because $$a^2\bmod 4 \in\{0,1\},$$ $$x\equiv y\equiv z \equiv 0\pmod4,$$ and that means impossibility of any solution with $\gcd(x,y)=1.$

So the original equation has no solution in rational numbers.

Yuri Negometyanov
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Transform your search for rational solutions into a search for positive integer solutions by setting $x=a/c$, $y=b/c$, so the equation becomes $$ a^2+b^2=3c^2 $$ Looking at this modulo $3$, we conclude that $a\equiv0\pmod{3}$ and $b\equiv{0}\pmod{3}$ (prove it). Therefore $a=3a_1$, $b=3b_1$ and so $$ 3a_1^2+3b_1^2=c^2 $$ which implies $c=3c_1$, giving $a_1^2+b_1^2=3c_1^2$.

Now we are done: assume a solution with positive $c$ exists and take one with minimal such $c$. We immediately get a contradiction.

egreg
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