$$ \log_2 \left(1 + \frac{1}{a}\right) + \log_2 \left(1 + \frac{1}{b}\right)+ \log_2 \left(1 + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.} $$

Apparently, the answer is $a= 1$, $b =2$, and $c\space = 3$.

When I asked my math teacher I was told that the solution involved a bit of number theory, but didn't recieve a complete explanation. Could someone clear that up for me?

Edit: I had made a mistake in typing the question. I had left it as:

$ \log_2 \left(a + \frac{1}{a}\right) + \log_2 \left(b + \frac{1}{b}\right)+ \log_2 \left(c + \frac{1}{c}\right) = 2 \quad \text{where $a$, $b$, $c \in N$.} $

My apologies for causing confusion.

  • 8
    $\begingroup$ Unless I am mistaken, the equation does not hold for $(a, b, c) = (1, 2, 3)$. Perhaps it is $\log_2 (1 + \frac{1}{a}) + \log_2 (1 + \frac{1}{b})+ \log_2 (1 + \frac{1}{c}) = 2$? In that case you can find a solution on AoPS: artofproblemsolving.com/community/c6h43200p273208. $\endgroup$
    – Martin R
    May 24, 2020 at 10:04
  • $\begingroup$ Yes, Indeed it is, I made a mistake. Thanks for pointing it out and sharing the link! $\endgroup$ May 24, 2020 at 10:56
  • 2
    $\begingroup$ After your edit, you put Ishan's answer, with a decent amount of upvotes, in an unfavorable light. Also, the title still points at the old version. $\endgroup$
    – rtybase
    May 24, 2020 at 10:57
  • $\begingroup$ @rtybase: On the other hand, if a question states that “apparently, the answer is ...” if it isn't then a request for clarification might be more appropriate than answering the (apparently wrong) question. $\endgroup$
    – Martin R
    May 24, 2020 at 11:11
  • $\begingroup$ @MartinR indeed, I was just highlighting the fact. I hope guys will find a way to sort this little problem out. I'd simply add notes in the question and answer mentioning the edit (and time of the edit) and the change in the meaning of the question. Or close this one and create another ... $\endgroup$
    – rtybase
    May 24, 2020 at 11:17

1 Answer 1


The problem, as you have written it, has no solution.

Simplifying the LHS, we get $$(a^2+1)(b^2+1)(c^2+1) = 4abc$$ But, by the AM-GM inequality, we get $x^2+1\ge2x$, which gives $$(a^2+1)(b^2+1)(c^2+1) \ge 8abc$$

  • $\begingroup$ I am so sorry. I made a mistake in typing out the question. Sincere apologies. $\endgroup$ May 24, 2020 at 10:53
  • 1
    $\begingroup$ @Apologies are fine...but not enough in this case. You have here a very nice answer and with clear explanations and you should leave the original question as it was, and then ask a new question ...! $\endgroup$
    – DonAntonio
    May 24, 2020 at 11:02
  • $\begingroup$ @DonAntonio True, I hadn't thought of doing that. But I don't want to create any more confusion by editing the question back to what it was. I will be sure to do what you have suggested in case I make a similar mistake in the future. $\endgroup$ May 24, 2020 at 11:27

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