Some notzero reals numbers are placed around circle. For every two neighbour numbers $a,b$ it true, that $a+b$ and $\frac{1}{a}+\frac{1}{b}$ are integer. Prove that there are not more than $4$ different numbers.
Problem
Source: St Petersburg Olympiad 2012, Grade 10, P4
Tags: number theory
WallyWalrus
09.10.2017 23:29
We say that a pair $(a,b)$ has the property $\mathcal{P}$ if $a,b\in\mathbb{R}\setminus\{0\}, a+b\in\mathbb{Z}, \dfrac{1}{a}+\dfrac{1}{b}\in\mathbb{Z}$. Some obviously properties: $(a,b)$ has the property $\mathcal{P} \Longleftrightarrow (-a,-b)$ has the property $\mathcal{P}$; $\forall a\in\mathbb{R}\setminus\{0\}, (a,-a)$ has the property $\mathcal{P}$. Let $a_1, a_2,\dots,a_n$ the numbers placed on the circle and $a\in\{a_1, a_2,\dots,a_n\}$. 1) If $a\in\mathbb{Q}\setminus\{0\}$, then $a_i\in\mathbb{Q}\setminus\{0\}, \forall i\in\{1,2,\dots,n\}$. 2) If $a\in\mathbb{R}\setminus\mathbb{Q}$, then $a_i\in\mathbb{R}\setminus\mathbb{Q}, \forall i\in\{1,2,\dots,n\}$. Let $(a,b)$ a pair with the property $\mathcal{P}$. Denote $a+b=S\in\mathbb{Z}, \dfrac{1}{a}+\dfrac{1}{b}=k\in\mathbb{Z}$. Case A). For $S=0$ results $b=-a$. Case B). $S\ne 0$. We will prove: if $a$ is a number on the circle, exists at most a single value $b\ne -a$ such that $(a,b)$ has the property $\mathcal{P}$. For $S\ne 0$ results $k\ne 0, ab=\dfrac{S}{k}$. Results: $a$ and $b$ are the solutions of the second degree equation $x^2-Sx+\dfrac{S}{k}=0$. $a,b=\dfrac{S\pm\sqrt{S^2-\dfrac{4S}{k}}}{2}$. B1). The rational solutions $(a,b)$ are $a=b, |a|\in\left\{\dfrac{1}{2};1;2\right\}$.
$a,b\in\mathbb{Q}\Longrightarrow \sqrt{S^2-\dfrac{4S}{k}}\in\mathbb{Q}$.
For $k=1, S=4$ results $a=b=2$.
For $k=1, S<0, S^2-2S+1<S^2-4S<S^2-4S+4\Longrightarrow (S-1)^2<S^2-4S<(S-2)^2\Longrightarrow \sqrt{S^2-4S}\notin\mathbb{Q}$.
For $k=1, S>4, S^2+2S+1<S^2+4S<S^2+4S+4\Longrightarrow (S+1)^2<S^2+4S<(S+2)^2\Longrightarrow \sqrt{S^2+4S}\notin\mathbb{Q}$.
For $k=-1, S=-4$ results $a=b=-2$.
For $k=2, S=2$ results $a=b=1$.
For $k=-2, S=-2$ results $a=b=-1$.
For $k=4, S=1$ results $a=b=\dfrac{1}{2}$.
For $k=-4, S=-1$ results $a=b=-\dfrac{1}{2}$.
For $k\in\{-1,\pm2,\pm4\}$ and the other values of $S$ results $\sqrt{S^2-\dfrac{4S}{k}} \notin\mathbb{Q}$.
For the other values of $k$ ($|k|=3$ or $|k|\ge 5$):
$S^2-\dfrac{4S}{k}=\dfrac{p^2}{q^2}, p,q\in\mathbb{N}, q\ne 0, gcd(p,q)=1$.
Let $gcd(|S|,|k|)=d\Longrightarrow |S|=d\cdot m, |k|= d\cdot l, d,m,l \in\mathbb{Z^+},gcd(m,l)=1, dq\ge 3$.
$S^2-\dfrac{4S}{k}=d^2m^2\pm\dfrac{4m}{l}=\dfrac{p^2}{q^2}\Longrightarrow l=q^2\Longrightarrow$
$\Longrightarrow d^2m^2q^2\pm4m=p^2\Longrightarrow (dq)^2m^2\pm4m=p^2$.
$(dq)^2m^2-2dqm+1<(dq)^2m^2-4m<(dq)^2m^2\Longrightarrow (dqm-1)^2<(dq)^2m^2-4m<(dqm)^2\Longrightarrow \sqrt{(dq)^2m^2-4m}\notin \mathbb{Z}$.
$(dq)^2m^2+2dqm+1>(dq)^2m^2+4m>(dq)^2m^2\Longrightarrow (dqm+1)^2>(dq)^2m^2+4m>(dqm)^2\Longrightarrow \sqrt{(dq)^2m^2+4m}\notin \mathbb{Z}$.
In the both cases, results $\sqrt{S^2-\dfrac{4S}{k}}\notin\mathbb{Q}$.
. Results: if exists $a\in\mathbb{Q}$ on the circle, there are maximum two distinct values on the circle: $a$ and $-a$, where $|a|\in\left\{\dfrac{1}{2};1;2\right\}$. 2) The irrational solutions $(a,b)$. If $\sqrt{S^2-\dfrac{4S}{k}}\notin\mathbb{Q}$, then $a=u+v\sqrt{t}, b=u-v\sqrt{t}$, where $u,v\in\mathbb{Q}\setminus\{0\}$ and $t\ge2$ is a square-free integer. This representation $a=u+v\sqrt{t}$ is unique (does not exist two different triplets $(u,v,t), (u_1,v_1,t_1)$ such that $a=u+v\sqrt{t}=u_1+v_1\sqrt{t_1})$.
a) For $t=t_1$:
$u-u_1=(v_1-v)\sqrt{t}$.
If $v\ne v_1\Longrightarrow \sqrt{t}=\dfrac{u-u_1}{v_1-v}\in\mathbb{Q}$, contradiction;
If $u\ne u_1\Longrightarrow \dfrac{1}{\sqrt{t}}=\dfrac{v_1-v}{u-u_1}\in\mathbb{Q}$, contradiction.
Results $(u,v,t)=(u_1,v_1,t_1)$.
b)For $t\ne t_1$:
$u+v\sqrt{t}=u_1+v_1\sqrt{t_1}\Longrightarrow uv_1-u_1v=vv_1(\sqrt{t_1}-\sqrt{t})\Longrightarrow \sqrt{t_1}-\sqrt{t}=\dfrac{uv_1-u_1v}{vv_1}=w\in\mathbb{Q}$.
$\sqrt{t_1}-\sqrt{t}=w\Longrightarrow t_1+t-2\sqrt{tt_1}=w^2\Longrightarrow \sqrt{tt_1}=\dfrac{t+t_1-w^2}{2}\in\mathbb{Q}$.
Let $t=p_1\cdot p_2\cdots p_I$ and $t_1=q_1\cdot q_2\cdots q_J$, where $p_1, p_2,\dots,p_I, q_1,q_2,\dots,q_J$ are prime numbers.
$\sqrt{tt_1}\in\mathbb{Q}\Longrightarrow t=t_1$, contradiction.
In this case, for a given $a=u+v\sqrt{t}$, exists a single value $b\ne -a$ such that $(a,b)$ has the property $\mathcal{P}$, respectively $b=u-v\sqrt{t}$. The possible neighbors of $a$ are $-a$ and $b$, the possible neighbors of $-a$ are $a$ and $-b$, the possible neighbors of $b$ are $-b$ and $a$, the possible neighbors of $-b$ are $b$ and $-a$. Conclusion: If $a$ is a number on the circle, $a\in\mathbb{R}\setminus\mathbb{Q}$, then $a$ has the form $a=u+v\sqrt{t}$ and the possible numbers on the circle are: $a_i\in\{u+v\sqrt{t}, u-v\sqrt{t}, -u+v\sqrt{t}, -u-v\sqrt{t}\}$. A possible configuration of the numbers on the circle is: $a,b,-b,-a$, in this order.