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Optic equation says for

$${\frac {1}{a}}+{\frac {1}{b}}={\frac {1}{c}}$$

All solutions in integers $a, b, c$ are given in terms of positive integer parameters $m, n, k$ by

$$a=km(m+n) , \quad b=kn(m+n), \quad c=kmn$$

where $m$ and $n$ are coprime.

How is this proved? I went through the reference, which says "A. Palmström, J. Sadier, and C. Moreau" "Ibid, 299-302". No clue what that means though.

Does anyone know how why all the solutions are covered by these formulas?

Bill Dubuque
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simonzack
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  • Ibid. is short for ibidem, Latin for "in the same place" (that is, same reference as previous line). – Théophile Oct 31 '20 at 14:10
  • I looked at the references, it appears that this is the correct reference for that result: Dickson, L. E., History of the Theory of Numbers, Volume II: Diophantine Analysis, Chelsea Publ. Co., 1952, pp. 688–691. – Derek Luna Oct 31 '20 at 14:10
  • @Théophile Ah that means it's L'Intermédiaire des math., 2, 1895, won't be in English though. – simonzack Oct 31 '20 at 14:19

2 Answers2

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The equation $$\frac 1a+\frac 1b=\frac 1 c$$ is equivalent to $$(a+b)c=ab\quad \text {or}\quad (a-c)(b-c)=c^2$$

Note that $(ka,kb,kc)$ is a solution iff $(a,b,c)$ is, so we may assume that $\gcd(a,b,c)=1$.

If $p$ is any prime dividing $c$ then $p$ must divide either $a-c$ or $b-c$. It can't divide both, else $p$ would divide $\gcd(a,b,c)$. Thus $p^2$ divides one of $(a-c),(b-c)$. It follows that both $(a-c)$ and $(b-c)$ are perfect squares, so we may write, e.g., $(a-c)=m^2,\,(b-c)=n^2$. Then, of course, $c=mn$.

Thus every relatively prime solution is given by $$a=m^2+mn=m(m+n)\quad \quad b=n^2+mn=n(m+n)\quad \quad c=mn$$

as desired.

lulu
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If there were no limits on the right side, then it would be

$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$.

Because of the limits on the right side, the denominator has to be divisible by $a+b$, thats why one multiplies this to the denominators. This would be enough, but for saying that $m$ and $n$ are coprime, its necessary to multiply the denominators by k.

buja
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