Show that for any $n\in\mathbb{N}$ there does not exist natural numbers $x,y$ such that $$\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}.$$


1 Answer 1


Hint: Assume on the contrary that there is such $x, y$. Show $x+y \geq 2n+1$. Write $x+y=2n+1+k, k \geq 0$ and show $(2n+1-k)^2-1<4xy<(2n+1-k)^2$.

Hint 2: To show the last part, square the original inequality, shift some terms, and square again. Proving the lower bound will require slightly more work than the upper bound.

Edit: Since you appear to have so much difficulty with the lower bound, I guess it would be best to explain in more detail.

As mentioned in the second hint, square both sides and subtract $x+y$ to get


Since $x+y=2n+1+k$,


The upper bound trivially implies $k<2n+1$, so $k \leq 2n$, so $k<2\sqrt{n}\sqrt{n+1}$.

Thus the lower bound is positive, and we may square again.


For the lower bound,

\begin{align} & (2\sqrt{n}\sqrt{n+1}-k)^2-((2n+1-k)^2-1)\\ & =(4n(n+1)+k^2-4k\sqrt{n}\sqrt{n+1})-(4n^2+4n+1+k^2-2k(2n+1)-1)\\ &=2k(2n+1-2\sqrt{n}\sqrt{n+1}) \\ &\geq 0 \end{align}

  • $\begingroup$ But how to show that $(2n+1-k)^2 -1 <4xy<(2n+1-k)^2 $? $\endgroup$
    – user110661
    Jun 4, 2014 at 15:52
  • $\begingroup$ @FisiaiLusia Have you shown $x+y \geq 2n+1$? What have you tried in relation to my hint? $\endgroup$
    – Ivan Loh
    Jun 4, 2014 at 15:54
  • $\begingroup$ Yes, if $x+y < 2n+1$ then $$\sqrt{x} +\sqrt{y} <\sqrt{2n+1 +2\sqrt{xy}} \leq \sqrt{n} +\sqrt{n+1} .$$ $\endgroup$
    – user110661
    Jun 4, 2014 at 16:04
  • $\begingroup$ Oh I see, if $x+y =2n+1+k $ and $\sqrt{x} +\sqrt{y} <\sqrt{4n+2}$ then $$2\sqrt{xy} <4n+2 -(x+y) =2n+1-k$$ hence $$4xy< (2n+1-k)^2$$ $\endgroup$
    – user110661
    Jun 4, 2014 at 16:10
  • $\begingroup$ But, how to show that $4xy>(2n+1-k)^2 -1$? $\endgroup$
    – user110661
    Jun 4, 2014 at 16:15

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