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I don't know where to start,

Find all integers $a$, $b$, $c$ that satisfy $a\sqrt{2}−b = c\sqrt{3}$.

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  • $\begingroup$ Hint: square both sides. Deduce that $\sqrt 2\in \mathbb Q$. $\endgroup$ – lulu Jul 2 '16 at 20:58
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    $\begingroup$ $a=b=c=0$ works. :-) $\endgroup$ – vadim123 Jul 2 '16 at 20:59
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    $\begingroup$ By the simple Lemma here we deduce that $\,1,\,\sqrt{2},\,\sqrt{3}\,$ are linearly independent over $\,\Bbb Q,\,$ since none of $\,\sqrt 2,\, \sqrt 3,\, \sqrt 6\,$ are in $ \Bbb Q.\,$ So the only solution is $\,a = b = c = 0.\ $ $\endgroup$ – Bill Dubuque Jul 2 '16 at 21:45
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    $\begingroup$ @lulu You need to say more than that, because squaring yields $\, 2ab\sqrt 2 \in \Bbb Q,\,$ but that doesn't imply $\,\sqrt 2 \in \Bbb Q\,$ if $\,a\,$ or $\,b = 0.\,$ But in those cases we can deduce $\,\sqrt3\,$ or $\,\sqrt 6\in \Bbb Q.\,$ The proof is done generally in the Lemma linked in my prior comment. $\endgroup$ – Bill Dubuque Jul 2 '16 at 22:07
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$1,\sqrt{2},\sqrt{3}$ are linearly independent: assuming that $$ a\sqrt{2}-c\sqrt{3} = b $$ we have $$ 2a^2+3c^2-b^2=2ac\sqrt{6} $$ but $\sqrt{6}\not\in\mathbb{Q}$.

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    $\begingroup$ You mean linearly independent, I’m sure. $\endgroup$ – Lubin Jul 2 '16 at 21:02
  • $\begingroup$ But vadim123's comment shows $a=b=c=0$ works. $\endgroup$ – Ahmed S. Attaalla Jul 2 '16 at 21:03
  • $\begingroup$ so, short of 0 , 0 , 0 you are saying that there are no integer solutions to a b and c. $\endgroup$ – DobbyIsDead Jul 2 '16 at 21:05
  • $\begingroup$ If $ac\neq 0$, then $\sqrt{6}=\frac{2a^2+3c^2-b^2}{2ac}\in\mathbb Q$, contradiction. Therefore $ac=0$, i.e. either $a=0$ or $c=0$. You'll get $a=b=c=0$ using the fact that $\sqrt{2}, \sqrt{3}$ are also irrational. $\endgroup$ – user236182 Jul 2 '16 at 21:44
  • $\begingroup$ @user236182 This type of argument can be done just as easy in general, e.g. see this Lemma. $\endgroup$ – Bill Dubuque Jul 2 '16 at 21:48

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