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# Good question!

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

Note by Kiran Patel
4 years, 6 months ago

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

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

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

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

$$k=-n$$ 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.

- 4 years, 6 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.

- 4 years, 6 months ago

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

- 4 years, 6 months ago

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.

- 4 years, 6 months ago

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

- 4 years, 6 months ago