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If $a,b,c$ are positive integers and $\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.

I was trying to get something useful from the information given in the question but was unable to get that.

Thomas Andrews
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Caratheodory_Enthusiast
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  • Ansatz: $a = c+d$. – Daniel Fischer Jun 27 '15 at 13:47
  • http://math.stackexchange.com/questions/1166999/find-all-integers-x-y-and-z-such-that-frac1x-frac1y-frac/1167475#1167475 http://math.stackexchange.com/questions/419766/number-of-solution-for-xy-yz-zx-n/708100#708100 http://math.stackexchange.com/questions/977926/dfrac1a-dfrac1b-dfrac1c-a-b-c-in-mathbbn-with-no-common-factor-fi/977944#977944 – individ Jun 27 '15 at 14:00

4 Answers4

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Let $d=(a,b)$ be the g.c.d. of $a$ and $b$. We will show that $a+b=d^2$. Write $a=a_1d$ and $b=b_1d$. You also have $(a_1,b_1)=1$. We thus have $$\frac{1}{a}+\frac{1}{b}=\frac{1}{d}\left(\frac{1}{a_1}+\frac{1}{b_1}\right)=\frac{1}{c}$$$$\frac{a_1+b_1}{a_1b_1}=\frac{d}{c}$$ Observing that $(d,c)=1$ and $(a_1+b_1,a_1b_1)=1$, we get that $a_1+b_1=d$ and $a_1b_1=c$. But then $$a+b=d(a_1+b_1)=d^2$$

Zain Patel
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  • One should say more than "we get" since the property that you implicitly use there is equivalent to uniqueness of prime factorizations. – Bill Dubuque Jun 30 '15 at 00:35
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$$ \frac1a+\frac1b=\frac1c\implies ac+bc=ab $$ $a\mid bc$. Therefore, each prime that divides $a$ must divide one and only one of $b$ or $c$ (since $(a,b,c)=1$). Therefore, $a=(a,b)(a,c)$. Similarly, $b=(a,b)(b,c)$ and $c=(a,c)(b,c)$.

Thus, $$ a+b=\frac{ab}c=\frac{(a,b)(a,c)(a,b)(b,c)}{(a,c)(b,c)}=(a,b)^2 $$

robjohn
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Ah well, since Jack deleted his answer, I'll fix it for him.

The equation is equivalent to $ac+bc=ab$ or $(a-c)(b-c)=c^2$.

Claim: $\gcd(a-c,b-c)=1$.

Proof: $\gcd(a,b,c)=1$ implies $\gcd(a-c,b-c,c)=1$ which implies $\gcd(a-c,b-c,c^2)=1$. But $a-c\mid c^2$, so $\gcd(a-c,b-c,c^2)=\gcd(a-c,b-c)$.

When the product of two relatively prime positive integers is a perfect square, the two numbers are perfect squares. So:

$$a-c=d^2$$ for some $d\mid c$, and $b-c=\frac{c^2}{d^2}$.

So $$a+b = c+d^2 + c+\frac{c^2}{d^2} = \left(d+\frac{c}{d}\right)^2$$

[ In particular, Jack's assumption that $d=1$ or $d=c$ was wrong because in the case:

$$\frac{1}{10} + \frac{1}{15}=\frac{1}{6}$$

we have $d=2$.]

Thomas Andrews
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Lemma $\ $ If $\ \color{#0a0}{(a,b,c,d)= 1}\ $ then $\ \color{#c00}{ab = cd}\,\Rightarrow\, d = (a,d)(b,d)$

Proof $\,\ (a,d)(b,d) = (\color{#c00}{ab},ad,bd,d^2) = (\color{#c00}{cd},(a,b,d)d) = \color{#0a0}{(c,a,b,d)}d = d\ \ \ $ QED

Thus if $\ (a,d) = n = (b,d)\ $ then $\ d = (a,d)(b,d) = n^2.\ $ This applies to OP since clearing denom's yields $\,ab = c(a\!+\!b)\,$ with $\ d = a\!+\!b\ $ so $\,(a,d) = (a,b) = (b,d).$

Bill Dubuque
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