Find all integers $n$ that can be written as $n = \textrm{lcm}(a, b) + \textrm{lcm}(b, c) + \textrm{lcm}(c, a)$ where $a$, $b$, $c$ are all integers.
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5
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4$\begingroup$ The title, besides being ungrammatical, makes no sense. $\endgroup$– Daniel AsimovCommented Apr 17 at 18:19
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2$\begingroup$ $(a,b,c) = (0,n,n)$ represents every non-negative integer $n$ if no strict positivity is required. $\endgroup$– ZeroxCommented Apr 17 at 20:46
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$\begingroup$ @Aeryk How do you get $4$? $\endgroup$– ZeroxCommented Apr 17 at 20:48
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$\begingroup$ Ignore my (soon to be deleted) comments, I wasn't checking carefully enough... $\endgroup$– AerykCommented Apr 17 at 20:50
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5$\begingroup$ Why did this question six upvotes although it completely lacks context and motivation. Couldn't this be even a (tricky) homework problem? $\endgroup$– Jochen WengenrothCommented Apr 18 at 13:25
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1 Answer
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I'm assuming $a,b,c$ are strictly positive, otherwise $(a,b,c)=(0,n,n)$ represents every non-negative integer $n$.
Claim: $n$ is representable iff $n \ne 2^k$ for all non-negative integer $k$.
Note that if $n$ is represented by $(a,b,c)$ ,then $kn$ is represented by $(ka,kb,kc)$ for every positive integer $k$. Since $(1,1,d) (d \ge 1)$ represents every odd number no less than $3$, when $n \ne 2^k$ it is representable.
Suppose there is some $n=2^k$ representable. Choose the least $k \ge 0$ - It's easy to see that in fact $k \ge 2$ since $\operatorname{lcm}(a,b) \ge 1$ for $a,b \ge 1$ - and assume that $(a,b,c)$ represents $n$. Since $n$ is even, there can not be more than $1$ odd number in $a,b,c$. If $a,b,c$ are all even, then $\left( \dfrac{a}{2},\dfrac{b}{2}, \dfrac{c}{2} \right)$ represents $\dfrac{n}{2} = 2^{k-1}$; If only one of $a,b,c$ is odd (say $c$), then $\left( \dfrac{a}{2},\dfrac{b}{2}, c \right)$ represents $\dfrac{n}{2} = 2^{k-1}$. Both contradict the minimality of $k$.
