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Find two positive integers $x$ and $y$ such

$$\sqrt{69+20\sqrt{11}}=\sqrt{x}+\sqrt{y}$$

I have worked intensively with this task but I really can't find a solution to this problem. I hope that I can get a hint here.

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  • $\begingroup$ Square both sides. You will have some terms with roots in and some terms without. Equate the terms without roots (linear equation in x and y). Equate the terms with roots - see if you can then reduce it to xy = 1100. Combined with the linear equation to give a quadratic for x and so on... $\endgroup$ – Paul Nov 10 '14 at 14:27
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$$ 69 + 20 \sqrt{ 11 } = x + y + 2\sqrt{ x y }$$ Can you develop a simultaneous set of equations?

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Squaring both sides yields $$\begin{array}{rrcl}&69+20\sqrt{11}&=&x+2\sqrt{xy}+y \\\implies &69+2\sqrt{1100}&=&(x+y)+2\sqrt{xy}\end{array}$$ If you compare both sides you can see that a good candidate is a pair $(x,y)$ such that $$\begin{cases}x+y=69 \\ xy=1100\end{cases}$$ From these equations you can deduce that $$x(69-x)=1100 \implies x^2-69x+1100=0 \implies(x-44)(x-25)=0$$ which gives $x=44$ or $x=25$. Substituting $x$ in the above equation yields two possible solutions $$(x,y)=(44,25) \qquad \text{ or } \qquad (x,y)=(25,44)$$

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  • $\begingroup$ First implication, $69 + 20\sqrt{11} = 69 + 2\sqrt{1100}$ or I didn't understand something... $\endgroup$ – Tacet Nov 10 '14 at 14:36
  • $\begingroup$ Och. I didn't notice missing zero, sorry. Now it makes sense. +1 $\endgroup$ – Tacet Nov 10 '14 at 14:44

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