# Find all integers $a,\,b,\,c$ that satisfy $a\sqrt2−b = c\sqrt3$. [closed]

I don't know where to start,

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

• Hint: square both sides. Deduce that $\sqrt 2\in \mathbb Q$. – lulu Jul 2 '16 at 20:58
• $a=b=c=0$ works. :-) – vadim123 Jul 2 '16 at 20:59
• 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.\$ – Bill Dubuque Jul 2 '16 at 21:45
• @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 2 '16 at 22:07

$1,\sqrt{2},\sqrt{3}$ are linearly independent: assuming that $$a\sqrt{2}-c\sqrt{3} = b$$ we have $$2a^2+3c^2-b^2=2ac\sqrt{6}$$ but $\sqrt{6}\not\in\mathbb{Q}$.
• But vadim123's comment shows $a=b=c=0$ works. – Ahmed S. Attaalla Jul 2 '16 at 21:03
• 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 2 '16 at 21:44