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This is a non-calculator question:

Which integers $a, b, c$ satisfy the equation $a \sqrt 2 − b = c \sqrt5$?

I've tried solving it through trial and error and the only solution I seem to be getting is $0,0,0$.

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  • $\begingroup$ Could you tell us where you found this problem? $\endgroup$ – A.P. Nov 20 '15 at 14:04
  • $\begingroup$ A practice admissions test for Maths undergrad $\endgroup$ – urviguglani Nov 20 '15 at 14:06
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    $\begingroup$ @RaziehNoori It isn't immediately obvious that $a\sqrt{2} - c \sqrt{5}$ is irrational. $\endgroup$ – A.P. Nov 20 '15 at 14:09
  • $\begingroup$ $\sqrt 2$ and $\sqrt 5$ are linearly independent over the field of rational $\mathbb Q$; hence the integers satisfying the equality are given by $(a,b,c)=(0,0,0)$ $\endgroup$ – Piquito Nov 20 '15 at 15:31
  • $\begingroup$ what's a field of rational Q? @Ataulfo $\endgroup$ – urviguglani Nov 20 '15 at 15:51
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Let us assume there are $a,b,c \in \mathbb{Z},ac \neq 0$ such that $a \sqrt 2 − b = c \sqrt5$

$$b=a\sqrt{2} -c\sqrt{5}$$ $$b^2=2a^2+5c^2-2ac\sqrt{10}$$ $$\sqrt{10}=\frac{2a^2+5c^2-b^2}{2ac}$$ $$\implies \sqrt{10} \space \text{is rational (contradiction)}$$

EDIT: As suggested by A.P., we will need to rule out the cases when $a=0$ and when $c=0$ for a complete solution.

When $a=0$,

$$\sqrt{5}=-\frac b{c} \space \Rightarrow\Leftarrow$$

When $c=0$,

$$\sqrt{2}=\frac b{a} \space \Rightarrow\Leftarrow$$

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    $\begingroup$ This isn't enough. You should note that $a = 0$ or $c = 0$ can't work, either. $\endgroup$ – A.P. Nov 20 '15 at 14:11
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    $\begingroup$ You want to assume that $ac\ne 0$... $\endgroup$ – user251257 Nov 20 '15 at 14:11
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    $\begingroup$ As I mentioned in my previous comment, this answer isn't complete because it doesn't rule out solutions of the form $(0,b,c)$ or $(a,b,0)$. It isn't hard to do so, but you should nonetheless. $\endgroup$ – A.P. Nov 20 '15 at 15:15
  • $\begingroup$ @A.P. thank you very much. $\endgroup$ – urviguglani Nov 20 '15 at 15:52
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Hint: square both sides of the equation and remember that $\sqrt2$ and $\sqrt5$ are irrational.

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