1. ## Pigeonhole principle

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.

2. 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

3. Originally Posted by rectril
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.
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.

4. Sorry BIG mistake
$\displaystyle \frac{ab+bc+ac}{abc}=1$

do not know how i missed that!
$\displaystyle 3a^2 + 2(B+C)a +BC= a^3+a^2B+a^2C+aBC\Rightarrow$

$\displaystyle 3a^2 + 2(B+C)a +BC= a^3+(B+C)a^2+aBC\Rightarrow$
$\displaystyle 3a^2 + 2(B+C)a +BC= a(a^2+(B+C)a+BC)$

I am working on it,
post you when i have something new

Sorry again

5. Originally Posted by awkward
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.
$\displaystyle a$ 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.

6. Originally Posted by rectril
$\displaystyle a$ 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.
That is good, but the problem says to find all solutions in positive integers.

So can you find more solutions? Or if not, can you prove the one you found is the only one?

7. can't be 1 since there are no and that can possibly satisfy the equation, so must be 2. So .
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