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> Prove that $x^2 + y^2 = 3$ has no rational solutions $(x,y) \in \mathbb Q^2$. Use this to prove that $x^2+y^2 = 3^k$ has no solutions $(x,y) \in \mathbb Q^2$ for each $k \in \mathbb {2N+1}$.
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I've looked at [this article](https://math.stackexchange.com/questions/473771/rational-solutions-for-x2y2-3) after trying the proof myself and it seems I've used the last answer only I'll modify the denominator because I don't understand why the denominator can be the same. Should it not be different like in the following?

So this clearly is a proof that calls for contradiction ;) 

Suppose that $x^2 + y^2 = 3$ has rational solutions $(x,y) \in \mathbb Q^2$.

Now you can write the equation as $(\dfrac ab)^2 + (\dfrac cd)^2 = 3$
$\qquad$ with $gcd(a,b,c,d)=1$

Then you can rewrite the expression in: $$\frac {a^2\cdot d^2 + b^2\cdot c^2}{b^2\cdot d^2} = 3$$ 

So now I'll make a parity check: 

$(i)$ if $a,b,c$ and $d$ are even $\Rightarrow$ this contradicts the assumption for which $gcd(a,b,c,d) =1$

$(ii)$ if $a,b$ and $c$ are even and $d$ is uneven it still makes the entire fraction even which will lead to $(i)$

$(iii)$ if $a$ is even and $b,c$ and $d$ are uneven $\Rightarrow$ 

Here I'll get stuck because I could write the entire fraction in terms of even and uneven expressions and the work it out but that seems too much work for this. 

$(iv)$ $a,b,c$ and $d$ are uneven 

This will give me the same problem as in $(iii)$.