Show that $\gcd(a,b)>1$

Question

Given are three natural numbers $a$, $b$ and $c$, for which

$$\frac1a+\frac1b=\frac1c,$$

show that $\gcd(a,b)>1$.

Could you someone provide a hint?

I already tried algebraic manipulation, but I just can't find a way to prove it...

Answer 1

We have $$ \frac{1}{c}=\frac{a+b}{ab}. $$ As $c=\dfrac{ab}{a+b}$, then $a+b>1$ is a divisor of $ab$, and hence there is a prime divisor $p$, such that $$ p\mid a+b \quad\text{and hence}\quad p\mid ab. $$ Now, if $p \,\big|\,ab$, then $p\mid a$ or $p\mid b$. Assume that $p\mid a$. (The case $p\mid b$ is dealt with in the same way.) But as $$ p\mid a \quad \&\quad p\mid a+b, $$ then $p\mid b$. Thus, $\,p\,\mid\, $gcd$(a,b)$, and hence $\,\,$gcd$(a,b)>1$.

Answer 2

We have $c<a$ and $c<b$. This is because $\frac{1}{a}<\frac{1}{c}$ (similarly, $\frac{1}{b}<\frac{1}{c}$). Simplifying the given expression, we also have $$c(a+b)=ab$$

If $\gcd(a,b)=1$, then both $a$ and $b$ do not divide $c(a+b)$. Why is that? What can you conclude from this?

Answer 3

This is because the formula looks like this. For the equation: $$\frac{1}{X}+\frac{1}{Y}=\frac{1}{A}$$

You can write a simple solution if the number on the decomposition factors as follows:
$$A=(k-t)(k+t)$$

then: $$X=2k(k+t)$$ $$Y=2k(k-t)$$ or: $$X=2t(k-t)$$ $$Y=-2t(k+t)$$

Answer 4

This is not an answer for your question. But you can use it for obtain an answer.
For any given positive integer $p,$ parametric solution of the Diophantine equation $$\frac{1}{p}=\frac{1}{x}+\frac{1}{y}$$ can be written in the form $x=ac(a+b),y=bc(a+b),$ where $p=abc.$

Proof
Let $$\frac{1}{p}=\frac{1}{x}+\frac{1}{y} ,x,y∈Z^+.$$
Then $x+y=t$ and $xy=pt$ for some $t∈Z^+.$
Now the quadratic equation $$z^2-tz+pt=0$$ has two integer roots $x,y.$
Discriminant of this equation can be written as $$Δ_(x,y)=t^2-4pt=q^2, q∈Z^+.$$
The quadratic equation $$t^2-4pt-q^2=0$$ gives the value of $t.$
$$Δ_t=16p^2+4q^2=4r^2,r∈Z^+.$$
$$4p^2+q^2=r^2,r∈Z^+.$$
This equation is of the form of Pythagoras equation.
Therefore $p=abc,q=(a^2-b^2 )c$ and $r=(a^2+b^2 )c$ where $a,b,c$ are parameters.
Backward substitution gives that $t=(a+b)^2 c.$
Hence we can obtain that
$$x=ac(a+b),y=bc(a+b).$$

Answer 5

Suppose to the contrary that $\gcd(a,b)=1$. Clearing fractions gives $bc+ac=ab$. We have $ac=ab-bc$, so $b|ac$. Since $\gcd(a,b)=1$, then $b|c$. So $b\le c$. But if you think about the equation $\frac1a+\frac1b=\frac1c$, for natural numbers $a,b,c$, you can conclude that $c$ should be the smallest of $a,b,c$.