# Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$ [closed]

Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $$a^2+b^2+c^2$$ is a square, where $$a,b,c\not=0$$ are rational numbers.

It can probably be solved by a quick factoring trick, but I really can’t figure it out.

Because $$a^2+b^2+c^2=a^2+b^2+\frac{a^2b^2}{(a+b)^2}=\frac{(a^2+ab+b^2)^2}{(a+b)^2}.$$ I used that: $$(a^2+ab+b^2)^2=a^4+2a^3b+3a^2b^2+2ab^3+b^4.$$

Below we show how to simply derive the value of $$\,\color{#90f}x,$$ using only $$\rm\color{#0a0}{\textit{difference of squares}}\,$$ factorization

\begin{align} a^2+b^2+\color{#c00}c^2\, &=\, \color{#90f}{x^2},\ \ \ {\rm let}\ \ \ X = (a+b) x\\ \smash[t]{\overset{\!\!\!\large \times\ (a+b)^2}\iff}\ (a+b)^2(a^2+b^2) + (\color{#c00}{ab})^2 &=\, X^2\ \ \ \ {\rm by}\ \ \ \left[\,(a+b)\color{#c00}c = \color{#c00}{ab}\,\right]^2\\[.3em] \iff\qquad\quad\ \ (a+b)^2(a^2+b^2) &=\, \color{#0a0}{X^2 - (\color{c00}{ab})^2}\ \\[.3em] \smash[b]{\iff\ \, (\underbrace{a^2+b^2+2ab}_{})(\underbrace{a^2+b^2}_{})} &= {(X-ab)(X+ab)}\\ X^{\phantom{|^{|}}}\!\!\!+ab\ \ \ \ \ \ \ \ \ X-ab,\ \!\! &\ \ \ X = a^2+b^2+ab \end{align}\qquad\qquad\quad\

Thus $$\ \color{#90f}x = \dfrac{X}{a+b} = \dfrac{a^2+b^2+ab}{a+b}\ = \dfrac{(a+b)^2-ab}{a+b} = a+b-c\$$ is a solution.

Remark  With that insight we can easily derive the key identity at the heart of it, which is simply the symmetric polynomial Newton identity $$\ p_2 = \color{#0a0}{p_1^2} - 2\, \color{#c00}{e_2},\$$ with $$\,C\to -C,\,$$ i.e.

$$A^2+B^2+C^2 = \color{#0a0}{(A+B-C)^2} + 2(\color{#c00}{(A+B)C-AB})\qquad$$

hence we infer that $$\ A^2+B^2+C^2\$$ is a $$\,\rm\color{#0a0}{square},\,$$ when $$\,\color{#c00}{(A+B)C = AB}$$.

Proof  expanding $$\,\color{#0a0}{((A+B)-C)^2}$$ yields RHS $$= (A+B)^2\!-2AB+C^2 =$$ LHS

• The denominator $\,a+b\neq 0,\,$ else $\,b=-a\,\Rightarrow\, \dfrac{1}c = \dfrac{1}a + \dfrac{1}{-a} = 0,\,$ contradiction. $\ \ \$ – Bill Dubuque Dec 25 '19 at 22:00
• One advantage of doing it this way is that it requires only simple mental arithmetic due to the use of the simple difference of squares factorization. Doing it other ways may be more difficult, e.g. it may require taking the square root of a quartic to find $\,x.\ \ \$ – Bill Dubuque Dec 26 '19 at 0:27

Let $$\dfrac1a=A$$ etc.

$$\implies A+B=C$$

$$a^2+b^2+c^2=\dfrac{(AB)^2+(BC)^2+(CA)^2}{(ABC)^2}$$

$$(AB)^2+(BC)^2+(CA)^2$$

$$=(AB)^2+C^2(A^2+B^2)$$

$$=(AB)^2+(A+B)^2(A^2+B^2)\text{ if }C=\pm(A+B)$$

$$=(AB)^2+(A^2+B^2+2AB)(A^2+B^2)$$ $$=(AB)^2+(A^2+B^2)^2+2\cdot AB\cdot(A^2+B^2)=?$$

Let $$\tilde c=-c$$. Therefore we have

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{\tilde c}=0\Rightarrow ab+a\tilde c+b\tilde c=0$$

Now we can use the following identity:

$$(a+b+\tilde c)^2=a^2+b^2+\tilde c^2+2\underbrace{(ab+a\tilde c+b\tilde c)}_{=0}=a^2+b^2+\tilde c^2$$

Therefore

$$a^2+b^2+c^2=a^2+b^2+ (-\tilde c)^2=a^2+b^2+\tilde c^2=(a+b+\tilde c)^2=(a+b-c)^2\Rightarrow$$

$$a^2+b^2+c^2=(a+b-c)^2$$

With the last equation we proved what we wanted.

So we have $$c(a+b)=ab$$. Now $$\begin{eqnarray}a^2+b^2+c^2 &=&\underbrace{a^2+\color{red}{2ab}+b^2}\color{red}{-2c(a+b)}+c^2\\ &=&(a+b)^2-2c(a+b)+c^2\\ &=& (a+b-c)^2 \end{eqnarray}$$

We exploit innate symmetry. The hypothesis is equivalent to $$\,\color{#c00}{(a+b)c = ab},\,$$ i.e. $$\,e_2(-c)=0\,$$ for $$\,e_2 = ab+bc+ca.\,$$ Since $$\,p_2 = a^2+b^2+c^2\,$$ is symmetric we can write it as a polynomial in the elementary symmetric polynomials $$\,e_i.\,$$ By Gauss's algorithm we subtract $$\,e_1^2\,$$ from $$\,p_2\,$$ yielding

$$p_2 - e_1^2 = a^2+b^2+c^2 - (a+b+c)^2 = -2(ab+bc+ca) = -2 e_2$$

which yields that $$\ p_2 = e_1^2 -2 e_2,\,$$ which for $$\,c\to -c\,$$ yields

$$a^2+b^2 + c^2\, =\, \underbrace{(a+b-c)^2}_{\large \rm square\ \color{#c00}{\Large \checkmark}} -2(\underbrace{\color{#c00}{ab-(a+b)c})}_{\large =\ \color{#c00}0}$$

Remark $$\ p_2 = e_1^2 - 2e_2 = e_1 p_1 - p_2\,$$ is a special case of Newton's Identities relating power sums and elementary symmetric polynomials. Since it is instructive to do so, we derived it from scratch using Gauss's constructive version of the Fundamental Theorem of Symmetric Polynomials.

Gauss's algorithm may be viewed as a special case of Gröbner basis methods (which may be viewed both as a multivariate generalization of the (Euclidean) polynomial division algorithm, as well as a nonlinear generalization of Gaussian elimination for linear systems of equation). Gauss's algorithm is the earliest known use of such a lexicographic order for term-rewriting (now mechanized by the Grobner basis algorithm and related methods).