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# Help me in this question

Find all positive integers $$x$$ and $$y$$ satisfying $$\dfrac1{\sqrt x} + \dfrac1{\sqrt y} = \dfrac1{\sqrt{20}}$$.

Note by Ved Sharda
1 year, 3 months ago

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ok. multiply both sides by$$\sqrt{20xy}$$ $\sqrt{20}(\sqrt{x}+\sqrt{y})=\sqrt{xy}$ change $$(\sqrt{x},\sqrt{y})\to (a,b)$$ $ab-\sqrt{20}(a+b)+20=20\\ (a-\sqrt{20})(b-\sqrt{20})=20$ notice how only $$a=k\sqrt{20},b=k_2\sqrt{20}$$ would yield positive integer answers. we check $(k-1)(k_2-1)20=20\Longrightarrow k=2,k_1=2\Longrightarrow x=80,y=80$ · 1 year, 3 months ago

I think that there is one more solution pair, namely $$(x,y) = (45,180)$$, as

$$\dfrac{1}{\sqrt{45}} + \dfrac{1}{\sqrt{180}} = \dfrac{1}{3\sqrt{5}} + \dfrac{1}{6\sqrt{5}} = \left(\dfrac{1}{3} + \dfrac{1}{6}\right)\dfrac{1}{\sqrt{5}} = \dfrac{1}{2\sqrt{5}} = \dfrac{1}{\sqrt{20}}.$$

My approach was to note that as $$\sqrt{20} = 2\sqrt{5}$$ we must have $$x = 5m^{2}$$ and $$y = 5n^{2}$$ for some positive integers $$m,n$$ both greater than $$2$$, yielding the equation

$$\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{2} \Longrightarrow 2m + 2n = mn \Longrightarrow mn - 2m - 2n = 0 \Longrightarrow (m - 2)(n - 2) = 4.$$

As $$4 = 2*2 = 1*4$$ we can either have $$(m,n) = (4,4)$$ or $$(m,n) = (3,6)$$, the former of which yields your solution and the latter my additional solution. · 1 year, 3 months ago

pls help me.If w is one of the root of x2+x+2.find w10+w5+3 · 1 year, 3 months ago