3

Conjecture:
Given integers $a>b>c>1$ such that $\gcd(a,b,c)=1$. Then all the positive integer solutions to $$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$$ is given by: $c$ any number and for each divisor $d<\sqrt{c}\:$ of $\,c\,$ with $\gcd(d,c/d)=1$ $$ \left\{ \begin{array}{l} a_d=c+d^2 \\ b_d=\frac{c}{d}\cdot(\frac{c}{d}+d)=c\cdot(c+d^2)/d^2 \end{array} \right. $$ It is easy to show that $a_d,b_d,c$ all are solutions, but how to prove they are the only solutions?

Lehs
  • 13,791
  • 4
  • 25
  • 77

1 Answers1

2

If $a$, $b$ and $c$ are such that $a>b>c>1$ amd $\gcd(a,b,c)=1$ and $\tfrac1a+\tfrac1b=\tfrac1c$, then aso $$bc+ac=ab.\tag{1}$$ This shows that $c$ divides $ab$, and similarly $a$ divides $bc$ and $b$ divides $ac$. Let $w=\gcd(a,b)$ and $v=\gcd(a,c)$ and $u=\gcd(b,c)$. Then $u$, $v$ and $w$ are pairwise coprime positive integers with $w>v>u$, and $$a=vw,\qquad b=uw,\qquad c=uv.$$ Plugging this into $(1)$ shows that $w=u+v$, and so every solution is of the form $$a=v(u+v),\qquad b=u(u+v),\qquad c=uv,$$ with $v>u>0$ coprime. Conversely, for any two coprime positive integers $v>u$ you have $$\frac{1}{v(u+v)}+\frac{1}{u(u+v)}=\frac{1}{uv}.$$ This shows that these are precisely all solutions.

This is equivalent to your characterization; take $d=u$ and $c=uv$.

Servaes
  • 63,261
  • 7
  • 75
  • 163
  • Feel free to explain the downvote. – Servaes Dec 02 '21 at 16:29
  • Interesting! Is this a special case of my conjecture? – Lehs Dec 02 '21 at 16:40
  • “It follows that…” How? Doesn’t seem immediately obvious to me. – Thomas Andrews Dec 02 '21 at 16:59
  • 2
    However, your equation can be seen as $$(a-c)(b-c)=c^2.$$ So yo need a factorization of $c^2$ into relatively prime factors, which is the same as factoring $c$ into relatively prime factors, $c=uv ,$ $u,v$ relatively prime, with $u<v.$ Then you get $b=u^2+uv,a=v^2+uv,$ which is your answer. – Thomas Andrews Dec 02 '21 at 17:04
  • @ThomasAndrews That was a bit of a leap indeed. I have added an intermediate step. I also like your approach. – Servaes Dec 02 '21 at 18:08
  • @Lehs It is a proof of your conjecture. – Servaes Dec 02 '21 at 18:09
  • 2
    Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Dec 02 '21 at 19:41
  • Glad you had time to answer before the question was closed. Thanks! – Lehs Dec 02 '21 at 20:05
  • Why a=gcd(a,b)gcd(a,c) ? If we multiply a by a factor relatively prime with u, nu and w, the gcd do not change, but the relation between a and the gcd's is not valid any longer. Am I missing smth? – Thomas Dec 04 '21 at 14:00
  • @Thomas Because $a$ divides $bc$ we have $a=\gcd(a,bc)$. Because $\gcd(a,b,c)=1$ we have $\gcd(a,bc)=\gcd(a,b)\gcd(a,c)$. – Servaes Dec 04 '21 at 14:03
  • I see you use the previous observation that a divides bc. Thanks! – Thomas Dec 04 '21 at 14:06