1. ## Natural number

So the problem is:
The numbers a, b and c fullfill, that a2+2b2=3c2. Show that (a+2b+3c)/(a+c)*((a+b)/(b+c)+(b-c)/b-a)) is a natural number.

I hope someone can help me!
thx

2. ## Re: Natural number

Is this the problem:

$Given:\ a,\ b,\ and\ c \in \mathbb R\ such\ that\ a^2 + 2b^2 = 3c^2,\ a \ne c,\ b \ne -\ c,\ and\ a \ne b;$

$Prove\ that\ \dfrac{a + 2b + 3c}{a - c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$

3. ## Re: Natural number

Originally Posted by JeffM
$Prove\ that\ \dfrac{a + 2b + 3c}{a - c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$
Typo Jeff: left denominator = a + c

That expression seems to always equal 4. Prove it does and we're done?

4. ## Re: Natural number

Will you show me how to prove it?

5. ## Re: Natural number

Not me!
But I'll tell you that by rearranging equation:
b^2 = (3c^2 - a^2) / 2
it is easy to show that all 3 variables are all even or all odd.
First 2 are:
(a,b,c) = (1,11,9) and (2,22,18).

But I'm sure that mon ami Jeffroi has figured out the proof
in his head and will ride in soon to save the day !

6. ## Re: Natural number

Thanks Denis for catching the typo.

$Given:\ a,\ b,\ and\ c \in \mathbb R\ such\ that\ a^2 + 2b^2 = 3c^2,\ a \ne -\ c,\ b \ne -\ c,\ and\ a \ne b;$

$Prove\ that\ \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) \in \mathbb N.$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{(b + a)(b - a)}{(b + c)(b - a)}+ \dfrac{(b - c)(b + c)}{(b - a)(b + c)} \right ) \implies$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{2b^2 - a^2 - c^2}{(b - a)(b + c)} \right ).$

$But\ a^2 + 2b^2 = 3c^2 \implies 2b^2 = 3c^2 - a^2.$

$\therefore \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{3c^2 - a^2 - a^2 - c^2}{(b - a)(b + c)} \right ) \implies$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{2(c^2 - a^2)}{(b - a)(b + c)} \right ) \implies$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(c - a)(a + 2b + 3c)}{(b - a)(b + c)} \implies$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(ac + 2bc + 3c^2 - a^2 - 2ab - 3ac)}{b^2 - ab + bc - ac} = \dfrac{2(3c^2 - a^2 - 2ab + 2bc - 2ac)}{b^2 - ab + bc - ac}.$

$But\ a^2 + 2b^2 = 3c^2 \implies 3c^2 = 2b^2 +a^2.$

$\therefore \dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = \dfrac{2(2b^2 + a^2 - a^2 - 2ab + 2bc - 2ac)}{b^2 - ab + bc - ac} = \dfrac{2 * 2(b^2 - ab + bc - ac)}{b^2 - ab + bc - ac} \implies$

$\dfrac{a + 2b + 3c}{a + c} * \left ( \dfrac{a + b}{b + c} + \dfrac{b - c}{b - a} \right) = 4 \in \mathbb N.\ QED$

Just lots of fussy algebra.

7. ## Re: Natural number

Nice wrapup! I gave up at end of my 1st sheet of equations!