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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 $k \in \mathbb {2N+1}$.

I've looked at this article 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 $(iii)$.

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You say "I don't understand why the denominator can be the same" but in effect you have written $x=ad/bd$ and $y=bc/cd$, having the same denominator. You do get $$(ad)^2+(bc)^2=3(bd)^2$$ so in effect $$u^2+v^2=3w^2\tag{*}$$ with $u=ad$ etc. You also fixate on evenness, that is divisibility by $2$. But in $(*)$ it's divisibility by $3$ that is important. The left side is divisible by $3$. But considering $u$ and $v$ modulo $3$ shows that this implies $u$ and $v$ have to be multiples of $3$. Then both sides of $(*)$ are multiples of $9$ so $w$ is also a multiple of $3$; so $3$ is a common factor of $u$, $v$, $w$. Take out this factor and repeat ad infinitum.

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  • $\begingroup$ Wow, what was I thinking. Come on that was very stupid of me. $\endgroup$ – Anonymous196 Oct 21 '17 at 18:59
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If each of $a/b$ and $c/d$ is in lowest terms, then $a^2/b^2$ and $c^2/d^2$ are also in lowest terms, and since their sum is an integer we must have $b^2=d^2$.

So we have $$ a^2+c^2 = 3b^2 $$ By working out the squares modulo $3$ we see that the only way for two squares to sum to a multiple of $3$ is that each of the squares is separately a multiple of $3$. This means that $3$ is a factor of both $a$ and $c$, so $a^2+c^2$ must be a multiple of $9$. But then $b$ must have a factor of $3$, contradicting that $a/b$ was in lowest terms!

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  • $\begingroup$ or $\dfrac cb$? $\endgroup$ – Anonymous196 Oct 22 '17 at 16:27
  • $\begingroup$ @AnonymousI yes, either will produce a contradiction. $\endgroup$ – Henning Makholm Oct 22 '17 at 17:09
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Suppose there are rational $x=a/b$ and $y=c/d$ that $x^2+y^2=3$. We can assume that $gcd(a,b)=gcd(c,d)=1$. Then, $$(ad)^2+(bc)^2 = 3(bd)^2$$ so $$ 3\mid(ad)^2+(bc)^2 \Longrightarrow 3|ad \;\wedge \; 3|bc $$ Then if $3|a$ we have $3|c$ so $3|bd$ and we have a contradiction.

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Suppose we have minimal solution in integers to the equation $A^2+B^2=3C^2$. Consider this equation modulo $3$, recall that the squares of $0,1,2$ are $0,1,1$ respectively, so $A$ and $B$ must be divisible by $3$. Let $A=3a$ and $B=3b$ giving $3a^2+3b^2=C^2 $ , so $C$ must be divisible by $3$, so we have $a^2+b^2=3c^2$, which contradicts the supposition of minimality.

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