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The sum of two distinct real number is a positive integer and the sum of their squares is 2. Compute the greater of these two real numbers. I tried to set up the equations first,:

$x^2 +y^2 =2\\ x+y = n,\enspace n \in \mathbb{N}$

Then I have no idea how to solve these two..

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$x + y = n, x^2+y^2 = 2 \Rightarrow 2 = (x+y)^2 - 2xy = n^2 - 2xy \Rightarrow n^2-2 = 2xy \leq \dfrac{(x+y)^2}{2} = \dfrac{n^2}{2} \Rightarrow n^2 \leq 4 \Rightarrow n \leq 2$. Can you continue ?

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  • $\begingroup$ You could even use $2xy < \frac{(x+y)^2}{2}$, as $x \neq y$. $\endgroup$ – Henno Brandsma Nov 16 '14 at 19:18

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