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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

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    Aren't only integer solutions allowed for diophantine equations? $\Rightarrow x,y \in \mathbb{Z}$ – 2017-02-27
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    We seek solutions in the set of rational numbers – 2017-02-27
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    @mrnovice the homogenous form of this equation is the diophantine equation. $a^2+b^2=3c^2$. – 2017-02-27
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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 $\pi\equiv3\pmod4$. – 2017-02-27

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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.

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    @skyking Fixed. Thanks. – 2017-02-27
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    Much better. I assume one should realize that there must be such a solution (with $\gcd(x,y)=1$) because else one can divide $x$, $y$ and $z$ and get another solution and repeat and eventually end up with $\gcd(x,y)=1$ – 2017-02-27
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    @skyking Of course, I mean these arguments. Thank you for detalization. – 2017-02-27
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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.