Let $a$ and $b$ be two positive integers such that $2ab$ divides $a^2+b^2-a$. Prove that $a$ is perfect square
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2016
Tags: number theory, Divisibility, FBH, Perfect Square
23.09.2018 01:50
If I'm correct, it even suffices to know that $ab$ divides $a^2+b^2-a$ (without the factor of $2$)...
23.09.2018 03:30
Tintarn wrote: But clearly also $r^2s^2 \mid a^2+b^2$ How did you get that...?
23.09.2018 08:38
By setting $d = GCD(a,b)$, we obtain $(a,b) = (sd,td)$, where $s$ and $t$ are two positive coprime integers. Consequently $(1) \;\; \frac{a^2 + b^2 - a}{2ab} = \frac{(sd)^2 + (td)^2 - sd}{2 \cdot (sd) \cdot (td)} = \frac{d(s^2 + t^2) - s}{2dst} \in \mathbb{N}$. Therefore $s | dt^2$ by (1), yielding $s | d$ since $GCD(s,t)=1$. Futhermore $d | s$ by (1), which means $d=s$ (since $s | d$). Hence $a = sd = s \cdot s = s^2$, i.e. $a$ is a perfect square. q.e.d.
23.09.2018 12:15
duck_master wrote: Tintarn wrote: But clearly also $r^2s^2 \mid a^2+b^2$ How did you get that...? Well, $rs \mid a$ and $rs\mid b$ so $r^2s^2 \mid a^2$ and $r^2s^2 \mid b^2$ so $r^2s^2 \mid a^2+b^2$...
18.04.2019 12:43
Can anyone tell whether this solution is correct or not.
21.04.2019 19:08
Kindly if anyone can tell whether my solution was correct or not.
21.04.2019 20:54
Similar to BWM 2013 P5 BWM 2013 P5 wrote: Let $m,n \in \mathbb{Z}^+$, such, $$mn | m^2+n^2+m$$Prove, $m$ is a perfect square Bosnia-Herzegovina Regional Olympiad 2016 Grade 10 P2 wrote: Let $a$ and $b$ be two positive integers such that $2ab$ divides $a^2+b^2-a$. Prove that $a$ is perfect square Solution: Let $\gcd (a,b)=d$ $\implies$ $\begin{cases} a=md \\\ b=nd \\\ \gcd (m,n)=1 \end{cases}$ $$\frac{a^2+b^2-a}{2ab} =\frac{m^2d^2+n^2d^2-md}{2mnd^2} =\frac{m^2d+n^2d-m}{2mnd} \in \mathbb{Z}$$Since, $m$ divides $2mnd$ $\implies$ $m ~ | ~ m^2d+n^2d-m $ but since, $\gcd (m,n)=1$ $\implies$ $m ~ | ~ d$ And, Since, $d$ divides $2mnd$ $\implies$ $d ~ | ~ m^2d$ $+n^2d-$ $m$ $\implies$ $d ~ | ~ m$ $$\implies m=d \implies a=md=m^2=d^2$$Hence done! $\qquad \blacksquare$