Given $\dfrac1a+\dfrac1b=\dfrac1c$, where $a, b, c \in \mathbb{N}$ with no common factor, find all solutions.

Actually, you can think this question as a follow up of this one. Today, I saw this question and thought whether such numbers really exits! And quickly I found some solutions: $(2,2,1), (3, 6, 2),$ $(4, 12, 3),$ $ (5, 20, 4),$ $(6, 30, 5), (7, 42, 6), (8, 56, 7), (9, 72, 8) \ldots$ (see my comments). Clearly, $(n+1, $ $n(n+1),$ $n)$ for $n \in \mathbb{N}$ is a solution. Interestingly these are the only solutions I found.

(Irrelevant after paw88789's comment) So, my question is: Is $(n+1, $ $n(n+1),$ $n)$ for $n \in \mathbb{N}$ the characterization of the above problem?

Update After paw88789's answer, I realize that there are solutions of other form also. So, I get back to my original question:

Find all solutions to the above problem.


marked as duplicate by Paul, Lost1, Claude Leibovici, drhab, HK Lee Oct 17 '14 at 11:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $\frac1a+\frac1b=\frac1c$ is the same as $bc+ac=ab$, or $c=\dfrac{ab}{a+b}$ $\endgroup$ – Martigan Oct 17 '14 at 8:06
  • $\begingroup$ math.stackexchange.com/questions/530915/… $\endgroup$ – Paul Oct 17 '14 at 8:09
  • 2
    $\begingroup$ How about $(10,15,6)$ as a solution of a different form? $\endgroup$ – paw88789 Oct 17 '14 at 8:10
  • $\begingroup$ Just to point out, I think it is better to write $\{a, b, c\} \subset \mathbb{N}$ rather than $a, b, c \in \mathbb{N}$. Although they both mean the same thing, I have seen the first option much more commonly. $\endgroup$ – user477343 Oct 15 '17 at 6:03

Fix $c$ and let $a=c+x, b=c+y$

Then you obtain $c(a+b)=ab$ or $2c^2+cx+cy=c^2+cx+cy+xy$ or $$c^2=xy$$

So for any $c$, if $x,y$ are coprime with $xy=c^2$ you will have $\frac 1{c+x}+\frac 1{c+y}=\frac 1c$ with the condition you require.

So take $c=12$ you could have $x=1, y=144$ with $\frac 1{13}+\frac 1{156}=\frac 1{12}$, or $x=9, y=16$ with $\frac 1{21}+\frac 1{28}=\frac 1{12}$

Now suppose $x,y$ are relatively prime with $xy=c^2$. Then $a+b=2c+x+y=(\sqrt x+\sqrt y)^2$ and with $x,y$ being coprime factors of a square number, they must both be squares.

Alternatively let the highest common factor of $a,b,c$ be $p$ and the highest common factor of $a,b$ be $pq$.

Then $\frac 1{pqr}+\frac 1{pqs}=\frac 1{pt}$ whence $t(r+s)=qrs$.

Now $r,s$ are coprime by construction so we must have $r+s|q$ so that $q=u(r+s)$ and $t=urs$. But $a,b,c$ would have the common factor $up$, hence $u=1$.

Whence the general solution is $$\frac 1{pr(r+s)}+\frac 1{ps(r+s)}=\frac 1{prs}$$.

Cancelling the common factor gives the solution Yiorgos has stated. And $r(r+s)+s(r+s)=(r+s)^2$


These are not all the triplets:

The solutions are: $\big(k(k+\ell),\ell(k+\ell),k\ell\big)$, for all $k,\ell\in\mathbb N$.

  • $\begingroup$ Yeah, true. I missed that! :-(. But you should mention $\gcd(k,l) =1$. $\endgroup$ – pushpen.paul Oct 17 '14 at 8:24

Constantly this question appears.

For the equation: $$\frac{1}{X}+\frac{1}{Y}=\frac{1}{A}$$ You can write a simple solution if the number on the decomposition factors as follows: $$A=(k-t)(k+t)$$ then: $$X=2k(k+t)$$ $$Y=2k(k-t)$$ or: $$X=2t(k-t)$$ $$Y=-2t(k+t)$$

Perhaps these formulas for someone too complicated. Then equation: $$XY+XZ+YZ=N$$ If we ask what ever number: $p$

That the following sum can always be factored: $p^2+N=ks$

Solutions can be written. $$X=p$$ $$Y=s-p$$ $$Z=k-p$$

These formulas on the forum more than once draw. Before you ask see the archive.

  • $\begingroup$ Please note: $X=2k(k+t)$, $Y=2k(k-t)$ or $X=2t(k-t)$,$Y=-2t(k+t)$ do not constitute all solutions since odd numbers are also possible. $\endgroup$ – pushpen.paul Oct 17 '14 at 8:52
  • $\begingroup$ @pushpen.paul look closely and check. These solutions give $A=k^2-t^2$ $\endgroup$ – individ Oct 17 '14 at 9:43

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