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..


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.