Good question!

Find all solutions in integers, of the equation $\frac{1}{m+n+k} = \frac{1}{k} + \frac{1}{m} + \frac{1}{n}$

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

$\large \frac{1}{m+n+k}-\frac{1}{k}=\frac{1}{m}+\frac{1}{n}$

$\large \frac{-(m+n)}{k(m+n+k)}=\frac{m+n}{mn}$

$-mn=k(m+n+k)$

$k^2+k(m+n)+mn=0$

Solving the quadratic gives us:

$k=-n$ or $k=-m$

An analogous argument gives us that $m=-n$ is also a solution.

My doubt is that whether we can have $m,n,k=0$ as solutions or not?

Although the equation actually satisfies, I am not sure.

Aditya Parson - 7 years, 11 months ago

If any of the variables were zero, then the fractions in the problem would be undefined. This the solutions are in the form (a, a, -a), where a is a real number not equal to zero.

Bob Krueger - 7 years, 11 months ago

But, what if we simplify the expression such that the variables are only in the numerator?

Aditya Parson - 7 years, 11 months ago

@Aditya Parson – We cannot clear out any factor or expression unless we know about the values it takes. Since we are given the expression, we will have to keep in mind that it is the base expression and solve accordingly.

Nishant Sharma - 7 years, 11 months ago

@Aditya Parson – It doesn't change the fact that the original expression would be undefined if the variables were 0.

Michael Tang - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$