MHB Is the Expression Involving Inverse Squares of Differences a Square?

[anemone]

Here is this week's POTW:

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Given that $a,\,b$ and $c$ are different real numbers.

Prove that the expression $\dfrac{1}{(a-b)^2}+\dfrac{1}{(b-c)^2}+\dfrac{1}{(c-a)^2}$ is a square.

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[anemone]

Congratulations to kaliprasad for his correct solution::)

Solution from kaliprasad:

Spoiler

If $x+y + z = 0\cdots(1)$
then $x+y=-z$ or $x^2+y^2+2xy = z^2\cdots(2)$
$\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2}$
= $\dfrac{y^2z^2+x^2z^2+x^2y^2}{(xyz)^2}$
= $\dfrac{z^2(y^2+x^2)+x^2y^2}{(xyz)^2}$
= $\dfrac{(x^2+y^2+2xy)(x^2+y^2)+x^2y^2}{(xyz)^2}$ from (2)
= $\dfrac{((x^2+y^2+xy)+xy)((x^2+y^2+xy)-xy)+x^2y^2}{(xyz)^2}$
= $\dfrac{(x^2+y^2+xy)^2 - x^2y^2 +x^2y^2}{(xyz)^2}$
= $\dfrac{(x^2+y^2+xy)^2}{(xyz)^2}$
= $(\dfrac{x^2+y^2+xy}{xyz})^2$

Now if we use $x= a-b,\,y = b-c,\,z = c-a$, we see that they satisfy the condition in (1).

Hence the given expression $\frac{1}{(a-b)^2} + \frac{1}{(b-c)^2} + \frac{1}{(c-a)^2}$ is a square.

Alternate Solution:

Spoiler

Note that the sum of the three positive reals as in our intended expression resulted in another non-negative real but we have the property that says every non-negative real number is a square.

We therefore have proved $\dfrac{1}{(a-b)^2}+\dfrac{1}{(b-c)^2}+\dfrac{1}{(c-a)^2}$ is a square.