Find all integer positive numbers $n$ such that: $n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers and $[p,q]$ represents the least common multiple of numbers $p,q$.
Problem
Source: Romanian JBTST IV 2007, problem 4
Tags: least common multiple, number theory proposed, number theory
13.05.2007 15:46
pohoatza wrote: Find all integer positive numbers $n$ such that: $n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers and $[p,q]$ represents the least common multiple of numbers $p,q$. Let $a=2^{p}q$, $b=2^{p}$ and $c=1$ Then $n=2^{p}(2q+1)$ And, since any positive integer n can be written as $n=2^{p}(2q+1)$, any positive integer may be written as $n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers. Sorry (edited later) : This demo does not work with $n=2^{p}$ since this would imply $a=0$. So it remains to study the case $n=2^{p}$ End (edited later than later) : It does not exist solution for $2^{p}$ : Let $a=2^{u}a'$, $b=2^{v}b'$ and $c=2^{w}c'$ with a', b' and c' odd and $u\geq v\geq w$ Then $n=2^{u}([a',b']+[a',c'])+2^{v}[b',c']=2^{v}(2^{u-v}([a',b']+[a',c'])+[b',c'])$ and, since $2^{u-v}([a',b']+[a',c'])+[b',c']$ is odd and $>1$, n can't be $2^{p}$ And the response is : "any $n>2$ different from $2^{p}$, $p>1$ -- Patrick
06.12.2009 18:59
pohoatza wrote: Find all integer positive numbers $ n$ such that: $ n = [a,b] + [b,c] + [c,a]$, where $ a,b,c$ are integer positive numbers and $ [p,q]$ represents the least common multiple of numbers $ p,q$. Lemma: $ n$ can be written on the form iff $ 2n$ can be written on that form. Proof: If $ n = [a,b] + [b,c] + [c,a]$ then $ 2n = [2a,2b] + [2b,2c] + [2c,2a]$. If $ 2n = [a,b] + [b,c] + [c,a]$ we have to cases: $ 2 \mid a,b,c$ then $ n = \left [ \frac {a}{2}, \frac {b}{2} \right ] + \left [ \frac {b}{2}, \frac {c}{2} \right ] + \left [ \frac {c}{2}, \frac {a}{2} \right ]$. $ 2 \mid a,b, 2 \nmid c$: $ n = \left [ \frac {a}{2}, \frac {b}{2} \right ] + \left [ \frac {b}{2}, c \right ] + \left [ c, \frac {a}{2} \right ]$. $ \square{}$ Now if $ n = 2m + 1$ then $ n = [m,1] + [1,1] + [1,m]$. But $ [a,b] + [b,c] + [c,a] \ge 3 > 1$ so $ n = 1$ has no solution. Using the lemma we obtain that $ n$ can be written on the form iff it is not a power of $ 2$. $ \square{}$