Not sure if this belongs here or in discrete.
Find all triples of positive integers a<b<c for which (1/a)+(1/b)+(1/c)=1 holds.
Have no idea how to get started on this. Thanks.
It given that $\displaystyle
a<b<c \quad \quad [1]
$
Why dont you get started by solving the equality:
$\displaystyle
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1 \Rightarrow \quad \quad [2]
$
$\displaystyle
\frac{a+b+c}{a\cdot b \cdot c}=1
$
or else
$\displaystyle
{a+b+c}={a\cdot b \cdot c}\quad \quad [3]
$
Next assume that:
$\displaystyle
b = a+B
$
$\displaystyle
c = a+C
$
By expanding the equation [3] we have that
$\displaystyle
3a+B+C = a^3+a^2B+a^2C+aBC
$
Since all are possitive integers we compare addition terms seperate:
$\displaystyle 3a<a^3$ for any $\displaystyle a>1$
$\displaystyle B<a^2B$ for any $\displaystyle a>1$
$\displaystyle C<a^2C$ for any $\displaystyle a>1$
So we conclude that
$\displaystyle
3a+B+C < a^3+a^2B+a^2C+aBC
$ for a>1
Finally if a=1 then
$\displaystyle
3+B+C = 1+B+C+BC \Rightarrow
BC=2
$
The last cannot hold because $\displaystyle 1<B<C$
So there is no such combination of positive integers that satisfy [1] and [2] conditions
Hint: The mention of the pigeonhole principle may be a red herring-- you may not need it here!
Can we have $\displaystyle a \geq 3$? If so, then $\displaystyle b \geq 4$ and $\displaystyle c \geq 5$. So $\displaystyle 1/a \leq 1/3$,
$\displaystyle 1/b \leq 1/4$, and $\displaystyle 1/c \leq 5$, hence $\displaystyle 1/a + 1/b + 1/c \leq 1/3 + 1/4 + 1/5$. Doesn't look good...
So we must have $\displaystyle a = 1$ or $\displaystyle a = 2$. Analyze these cases separately and see where they lead you.
Sorry BIG mistake
[2] leads at:
$\displaystyle
\frac{ab+bc+ac}{abc}=1
$
do not know how i missed that!
then this leads to:
$\displaystyle
3a^2 + 2(B+C)a +BC= a^3+a^2B+a^2C+aBC\Rightarrow
$
$\displaystyle
3a^2 + 2(B+C)a +BC= a^3+(B+C)a^2+aBC\Rightarrow
$
$\displaystyle
3a^2 + 2(B+C)a +BC= a(a^2+(B+C)a+BC)
$
I am working on it,
post you when i have something new
Sorry again
$\displaystyle a$ can't be 1 since there are no $\displaystyle b$ and $\displaystyle c$ that can possibly satisfy the equation, so $\displaystyle a$ must be 2. So $\displaystyle 1/b + 1/c = 1/2$. Choose $\displaystyle b = 3$ and solve for c. $\displaystyle a = 2, b = 3, c = 6$ satisfy the equation.
Since $\displaystyle a=2$ we have that $\displaystyle b \geq 3$.
Following awkwards' procedure, we check if there's an upper limit for $\displaystyle b$, and there is indeed. If $\displaystyle b \geq 4$ then $\displaystyle c \geq 5$ and we have that $\displaystyle 1/b + 1/c \leq 1/4 + 1/5 < 1/2$
So $\displaystyle b \leq 3$ and since $\displaystyle b \geq 3$, we have that $\displaystyle b = 3$ which rectril has solved for already