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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Hint: square both sides. Deduce that $\sqrt 2\in \mathbb Q$. – lulu Jul 02 '16 at 20:58
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2$a=b=c=0$ works. :-) – vadim123 Jul 02 '16 at 20:59
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1By 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.\ $ – Bill Dubuque Jul 02 '16 at 21:45
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1@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. – Bill Dubuque Jul 02 '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 $\sqrt{6}\not\in\mathbb{Q}$.
Jack D'Aurizio
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so, short of 0 , 0 , 0 you are saying that there are no integer solutions to a b and c. – DobbyIsDead Jul 02 '16 at 21:05
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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. – user236182 Jul 02 '16 at 21:44
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@user236182 This type of argument can be done just as easy in general, e.g. see this Lemma. – Bill Dubuque Jul 02 '16 at 21:48
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