Olympiad Maths Prep

Library / /14 of 45

Algebra Difficulty 5.6 AIME, harder Prove it Ukraine

The positive integers xx, yy satisfy the conditions:
{x2+2y}>23,{y2+2x}>23. \{\sqrt{x^2 + 2y}\} > \frac{2}{3}, \{\sqrt{y^2 + 2x}\} > \frac{2}{3}.
Prove that x=yx = y.
Here, {a}∈[0;1)\{a\} \in [0; 1) denotes the fractional part of the number aa, that is, there exists an integer nn for which the equality a=n+{a}a = n + \{a\} holds. For example, {3.14}=0.14\{3.14\} = 0.14.

Solution

Suppose that for some positive integers x<yx < y these inequalities are true: {x2+2y}>23,{y2+2x}>23\{\sqrt{x^2 + 2y}\} > \frac{2}{3}, \{\sqrt{y^2 + 2x}\} > \frac{2}{3}. Note that y2<y2+2x<(y+1)2y^2 < y^2 + 2x < (y + 1)^2, so we have
y2+2x>(y+23)2⇔2x>43y+49⇒x>23y. y^2 + 2x > \left(y + \frac{2}{3}\right)^2 \Leftrightarrow 2x > \frac{4}{3}y + \frac{4}{9} \Rightarrow x > \frac{2}{3}y.

x2+2y>(x+1+23)2⇔2y>103x+259⇒y>53x. x^2 + 2y > \left(x + 1 + \frac{2}{3}\right)^2 \Leftrightarrow 2y > \frac{10}{3}x + \frac{25}{9} \Rightarrow y > \frac{5}{3}x.
But then, xy>xy⋅23⋅53xy > xy \cdot \frac{2}{3} \cdot \frac{5}{3}, a contradiction that completes the proof.

Looking for a route rather than an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.