There are given two positive real number $a<b$. Show that there exist positive integers $p, \ q, \ r, \ s$ satisfying following conditions: $1$. $a< \frac{p}{q} < \frac{r}{s} < b$. $2.$ $p^2+q^2=r^2+s^2$.
Problem
Source: Polish Mathematical Olympiad 2016 P5- Final Round
Tags: algebra, real number, Sum of Squares
j___d
12.07.2016 02:00
Let $x, y$ be any rational numbers from the interval $(0,1)$. Then the positive numbers
$1)\begin{array}{cccc}u=\dfrac{1-x^2}{1+x^2} & v=\dfrac{2x}{1+x^2} & w=\dfrac{1-y^2}{1+y^2} & z=\dfrac{2y}{1+y^2}\end{array}$
are also rational, moreover, $u^2+v^2=1$ and $w^2+z^2=1$. Let us write these four rational numbers as fractions with a common denominator:
$2)\begin{array}{cccc}u=\dfrac{q}{D} & v=\dfrac{p}{D} & w=\dfrac{s}{D} & z=\dfrac{r}{D}\end{array}$.
We see that $p^2+q^2=r^2+s^2=D^2$, which means that the natural numbers $p, q, r, s$ satisfy condition 2.
Now we prove $a<\dfrac{p}{q}<\dfrac{r}{s}<b$, which can be rephrased as
$a<\dfrac{2x}{1-x^2}<\dfrac{2y}{1-y^2}<b$
The middle equality holds when $x<y$. The other two, $ax^2+2x-a>0$ and $by^2+2y-b<0$, hold when $x>\dfrac{\sqrt{1+a^2}-1}{a}=:A$ and $y<\dfrac{\sqrt{1+b^2}-1}{b}=:B$.
It is clear that $A>0$ and $B<1$. Moreover, $A<B$, because
$A=\dfrac{\sqrt{1+a^2}-1}{a}=\dfrac{a}{\sqrt{1+a^2}+1}=\dfrac{1}{\sqrt{\dfrac{1}{a^2}+1}+\dfrac{1}{a}}<\dfrac{1}{\sqrt{\dfrac{1}{b^2}+1}+\dfrac{1}{b}}=B$.
The interval $(A,B)$ is therefore non empty and contained in $(0,1)$.
This means there exist infinitely many pairs $(\dfrac{p}{q},\dfrac{r}{s})$ that satisfy both conditions. To generate one, pick any $x, y$ that $A<x<y<B$, calculate $u, v, w, z$ using formulas 1) and rephrase them as 2).
Vexation
12.08.2016 06:12
Observation : Suppose $\frac{p}{q} < \frac{r}{s}$ and $p^2 + q^2 = r^2 + s^2$. Then, for $a,b,c,d$ s.t. $ab=cd$, $\frac{ap+br}{aq+bs}$ and $\frac{cp+dr}{cq+ds}$ divides $\frac{p}{q}$ and $\frac{r}{s}$ in the ratio of $a:b$ and $c:d$ respectively, and $(ap+br)^2 + (aq+bs)^2 = (cp+dr)^2 + (cq+ds)^2$. Take integer $n$ and rational $x$ s.t. $\frac{1}{n} < a <x< b < n$. Suppose $x$ divides $\frac {1}{n}$ and $n$ in the ratio of $y:z$. Numbers that divide $\frac{1}{n}$ and $n$ in the ratio of $ym : z(m+1)$ and $y(m+1):zm$ converge to $x$ when $m$ diverges, and satisfy condition #2 by the observation.
Pathological
24.02.2019 22:02
Alternatively, just take a big $n \in \mathbb{N}$ and consider all complex numbers of the form $(3+4i)^i * 5^{n-i}$. Since the argument of $3+4i$ isn't a rational multiple of $\pi$, we know that there exists two $i$ such that their arguments are both within $(\arctan{a}, \arctan{b})$ for sufficiently large $n$. Taking $p, q, r, s$ to be the real and complex parts of these two complex numbers accordingly finishes.
amogususususus
01.06.2024 08:03
I'm sorry if there is a mistake in my solution, not quite sure if this works. Construct a cyclic quadrilateral $ABCD$ with $\angle ADC = \angle ABC=90^{\circ}$. Note that setting $AD=q$, $CD=q$ and $AB=s$, $BC=r$ satisfies the second condition. Note that $$\frac{p}{q} = tan \ \angle ACD$$and $$\frac{r}{s} = tan \ \angle ACB$$With the restriction $0^{\circ} < \angle ACD, \angle ACB < \frac{\pi}{2}$. Let $\angle ACD=x$ and $\angle ACB=y$. Note that the function $f(t)= tan \ t$ is surjective over $\mathbb{R^+}$ for $t \in \left(0^{\circ},\frac{\pi}{2}\right)$. By Generalized Rational Density Theorem, we can find $x,y \in \left(0^{\circ},\frac{\pi}{2}\right)$ that satisfy $$a<tan \ x < tan \ y < b$$Hence we are finished.