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I'm trying to show that $(2+\sqrt{2})(3+\sqrt{3})$ is not a square in $F= \mathbb{Q}(\sqrt{2},\sqrt{3})$,following the method given by the answer in the post:Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $.
We have $a = (2+\sqrt{2})(3+\sqrt{3})$. Suppose that $a = c^2$ for some $c \in \mathbb{Q}(\sqrt{2},\sqrt{3})$. We write $c = v+w\sqrt{3}$ for some $v,w \in \mathbb{Q}(\sqrt{2})$. We then have \begin{align*} (v+w\sqrt{3})^2 &= (2+\sqrt{2})(3+\sqrt{3})\\ v^2+3w^2 + 2\sqrt{3}vw &= 6+2\sqrt{3} + 3\sqrt{2} +\sqrt{2}\sqrt{3}\\ (v^2+3w^2) +(2vw)\sqrt{3} &= 6+2\sqrt{3} + 3\sqrt{2} +\sqrt{2}\sqrt{3}\\ (v^2+3w^2) +(2vw) \sqrt{3} &= 3(2+\sqrt{2}) +(2+\sqrt{2})\sqrt{3}. \end{align*} From here, we get $v^2+3w^2 = 3(2+\sqrt{2})$ and $2vw = 2+\sqrt{2}$. And $2vw = 2+\sqrt{2}$ implies $w = \frac{2+\sqrt{2}}{2v}$. We plug $w = \frac{2+\sqrt{2}}{2v}$ into the equation $v^2+3w^2 = 3(2+\sqrt{2})$. We get \begin{align*} v^2 +3(\frac{2+\sqrt{2}}{2v})^2 &= 3(2+\sqrt{2})\\ v^2 +3(\frac{2+\sqrt{2}}{2v})^2 -3(2+\sqrt{2}) &= 0 \\ v^2+ \frac{3(2+\sqrt{2})^2}{4v^2} -3(2+\sqrt{2}) &= 0\\ v^4 +\frac{3}{4} (2+\sqrt{2})^2 - 3(2+\sqrt{2})v^2 &= 0\\ v^4 - 3(2+\sqrt{2}) v^2 + \frac{3}{4} (2+\sqrt{2})^2 &= 0\\ v^4 - 3(2+\sqrt{2})v^2 +[\frac{3}{2}(2+\sqrt{2})]^2 - [\frac{3}{2}(2+\sqrt{2})]^2 + \frac{3}{4} (2+\sqrt{2})^2 &= 0\\ v^4 - 3(2+\sqrt{2})v^2+ \frac{9}{4}(2+\sqrt{2})^2 - \frac{9}{4}(2+\sqrt{2})^2 + \frac{3}{4} (2+\sqrt{2})^2 &= 0\\ (v^2-\frac{3}{2}(2+\sqrt{2}))^2 &= \frac{9}{4} (2+\sqrt{2})^2 - \frac{3}{4}(2+\sqrt{2})^2. \end{align*} We then get \begin{align} (v^2 - \frac{3}{2}(2+\sqrt{2}))^2 &= \frac{3}{2}(2+\sqrt{2})^2\\ &= \frac{3}{2}(4+2+4\sqrt{2})\\ &= \frac{3}{2}(6+4\sqrt{2})\\ &= 9+6\sqrt{2}. \end{align} We write $v^2 -\frac{3}{2}(2+\sqrt{2})$ as $a+b \sqrt{2}$ with $a,b \in \mathbb{Q}$. And we get $a^2 + b^2 = 9$, $2ab = 6$. And I don't think this could give me any contradiction. I'm wondering if there's any mistake in my proof. Please help! Thanks!

Bill Dubuque
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Miranda
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    Why don't you use the simple norm method described in the linked question (not in its answer) . This is the same method that you were given as a hint in your prior deleted question. I closed your prior question with that dupe target because I thought the norm method would be clearer as described there. Do you not understand it? – Bill Dubuque Apr 02 '25 at 21:03
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    Note: as usual, I intended to add a closing comment to you prior question. explaining how the dupe applies. But your very quick deletion prevented that. Worth strong emphasis: you should strive to understand the norm method (not ad-hoc methods like above) since such methods are keystones of number theory and Galois theory, e..g. see this answer. $\ \ $ – Bill Dubuque Apr 02 '25 at 21:24
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    I have adjusted the tags to match your question, viz. why $a^2 + b^2 = 9$, $2ab = 6$ has no rational solutions. $\ \ $ – Bill Dubuque Apr 02 '25 at 21:48

2 Answers2

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HINT:

$$a^2+b^2=9; 2ab = 6 \implies a^2 +2ab+b^2 = 15 $$

$$\implies (a+b)^2=15. $$

With $a,b$ rational that is....

Mike
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  • In case additional clarification is needed, if an integer $x$ is not a square of another integer [such as $x=15$, indeed note that the integer $15$ is not a square of any other integer], then $x$ is also not the square of any rational number either. So $x=15$ is not the square of any other rational, and thus, there is no rational number $c$ such that $c^2=15$. And thus, there are no rational numbers $a$ and $b$ such that $(a+b)^2=15$ [for $c=a+b$ is rational if $a$ and $b$ both are]. – Mike Apr 03 '25 at 21:28
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From $2ab=6$ we get $b={3\over a}$. Substituting into $a^2+b^2=9$ we get

$$a^2+{9\over a^2}=9$$ $$a^4-9a^2+9=0$$

This is a quadratic equation in $a^2$, and the discriminant isn't a square (of a rational number), thus $a^2$ must be irrational, as well as $a$ itself.

lisyarus
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