$\sqrt{a+b\sqrt{c}}=\sqrt{x}+\sqrt{y}$
where $a, b, c\in\mathbb{Z}^+$ and x, y $\in \mathbb{Q} $
Please help show how to disprove or prove. Thanks a lot
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$\begingroup$ Could you perhaps provide some context for this problem, and explain what you have tried so far? This makes it easier for us to provide helpful answers. Also, can you clarify what you mean by "any unique $a,b,c \in \mathbb{Z}^+$"? Do you mean that $a,b,$ and $c$ are all different? $\endgroup$ – vociferous_rutabaga Jul 21 '14 at 17:27
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$\begingroup$ yeah, i was just helping a student a math, I solve that but i wonder though $\endgroup$ – user145472 Jul 21 '14 at 17:31
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$\begingroup$ the question was $\sqrt{16+4\sqrt{15}}=\sqrt{a}+\sqrt{b}$ some algebraic work yield 6 and 10 but i wonder if i change 16, 4 and 15, can i still yield integral result if i put any arbitary integers? i guess not likely. but I guess is it possible to yield rational number result. I guess probably? $\endgroup$ – user145472 Jul 21 '14 at 17:32
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Sorry about my previous answer, hopefully I've learned to read questions properly.
In the case where $c$ is not a square number there is a criterion.
We have:
$$ \sqrt{a+b\sqrt{c}} = \sqrt{x} + \sqrt{y} \iff a +b\sqrt{c} = x+y+ 2\sqrt{xy} $$
As $c$ is not square then this holds if and only if:
$$ x+y =a \ , \ \sqrt{xy} = \frac{b \sqrt{c}}{2} $$
Now squaring the second equation:
$$ x+y =a \ , \ xy = \frac{b^2 c}{4} $$
Now consider the quadratic equation in $z$ : $(z-x)(z-y) = 0 $.
Expanding this equation gives: $ z^2 - a z + \frac{b^2 c}{4} = 0 $
Applying the quadratic formula gives:
$$ x,y \in \lbrace \frac{a \pm \sqrt{a^2-b^2c} }{2} \rbrace $$
and therefore in this case a solution exists iff $a^2 - b^2c = n^2 $
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$\begingroup$ this answer is wrong, i didn't see that a,b,c were integers $\endgroup$ – JC574 Jul 21 '14 at 17:47
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$\begingroup$ There definitely exists a solution when $a^2 -b^2 c $ is square, you can get that from squaring the equation and setting x+y = a. for example in your case 16^2 - 15*4^2 = 16 = 4^2 , so there is a solution. $\endgroup$ – JC574 Jul 21 '14 at 17:52
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$\begingroup$ how about $\sqrt{12+4\sqrt{3}}$, i make it up casually but i can't solve it similarly $\endgroup$ – user145472 Jul 21 '14 at 18:04
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$\begingroup$ have you seen my edit to my answer? we can use the criterion there on that case: 3 is not square, and 12^2 - 3*4^2 = 96 is not square, so no solution. $\endgroup$ – JC574 Jul 21 '14 at 18:05
