Let $a_1$, $b_1$, $c_1$ be natural numbers. We define:

$$a_2 = (b_1, c_1), b_2 = (c_1, a_1), c_2 = (a_1, b_1)$$and $$a_3 = [b_2, c_2], b_3 = [c_2, a_2], c_3 = [a_2, b_2]$$ Show that $(b_3, c_3) = a_2$.

(due credits: Regional Mathematics Olympiad 2013 examination)


Substituting values, we have: $(b_3,c_3)=([c_2,a_2],[b_2,a_2])$

Now, we know that: $([m,n],[n,p])=[(m,p),n]$

So, $([c_2,a_2],[b_2,a_2])=[(c_2,b_2),a_2]=[((a_1,b_1),(c_1,a_1)),a_2]$

Now, we know that $((m,n),(n,r)) = (m,n,r)$.

So, $[((a_1,b_1),(c_1,a_1)),a_2] = [(a_1,b_1,c_1),a_2]=[(a_1,b_1,c_1),(b_1,c_1)]$

Thus, we have: $$(b_3,c_3)=[(a_1,b_1,c_1),(b_1,c_1)]$$

But, I am stuck here. I feel pretty close to the answer. But can't figure out what to do next. Perhaps, there's some identity related to $[(m,n),n]$?

UPDATE: I know that a solution exists at this site. However, I want to know how I can extend my solution to get the answer.

  • $\begingroup$ @NormalHuman Edit done. $\endgroup$ – Gaurang Tandon Nov 21 '15 at 12:29
  • $\begingroup$ Hint due you know principles of symmetry. $\endgroup$ – Archis Welankar Nov 21 '15 at 13:07
  • $\begingroup$ @ArchisWelankar No I don't. Could you explain that principle in an answer please? $\endgroup$ – Gaurang Tandon Nov 21 '15 at 16:03
  • $\begingroup$ Its an observatory principle not a theorem to be proved . $\endgroup$ – Archis Welankar Nov 21 '15 at 16:09
  • $\begingroup$ @ArchisWelankar Of course, but I need to know what it is to be able to conclude something about it, right? Could you please tell me at least what it is, so that I can apply it in the question. $\endgroup$ – Gaurang Tandon Nov 21 '15 at 16:40

It is clear that $$(a_1,b_1,c_1) | (b_1,c_1)$$ and if $m|n$ then $[m,n] = n$ so $$(b_3,c_3) = [(a_1,b_1,c_1),(b_1,c_1)] = (b_1,c_1) = a_2$$

  • $\begingroup$ Wow! It was so easy. Thanks! $\endgroup$ – Gaurang Tandon Nov 22 '15 at 11:10

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