<s>[INST] diophantine equation $x^2+y^2=3$ 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
 [/INST] 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.
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