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Find all pairs of positive whole numbers x and y which are a solution for $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $.

I don't really understand how to tackle this question. I rewrote $ \dfrac{2}{x} + \dfrac {3}{y} = 1 $ as $2y+ 3x =xy$ but that's it..

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If you multiply your equation with $x$ and $y$, you end up with $2y+3x = xy$ – Stefan Feb 14 '13 at 19:40
Yep, I made a little mistake, corrected! – JohnPhteven Feb 14 '13 at 19:42
up vote 2 down vote accepted

You can write the equation as $xy-2y-3x=0$. If you factor this you will get \begin{equation*} (x-2)(y-3)=6. \end{equation*} Do you see how to proceed?

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You could manually fill in x's and y's which satisfy the equality, but I wouldn't know how to do it otherwise – JohnPhteven Feb 14 '13 at 19:47
That's the way to do it. Note that $6=1\cdot 6=2\cdot 3$. One possible solution is thus given by $x-2=1$ and $y-3=6$. The other ones are found similarly. – user60725 Feb 14 '13 at 19:51
I see, thank you – JohnPhteven Feb 14 '13 at 19:58

If we multiply both sides the original equation by $xy$, we ge $2y+3x=xy$. We can rewrite this as $2y+3x-xy=0$. We now perform a little trick.

Note that what we have is very much like $(3-y)(x-2)=3x+2y-xy-6$. If we subtract $6$ from both sides, we get $2y+3x-xy-6=-6$ and $(3-y)(x-2)=-6$. Multiplying through by $-1$ gives $(y-3)(x-2)=6$. You can now search for pairs of small $x$ and $y$ that satisfy the condition. (Hint: Use the prime factorization of $6$ to help you.)

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The "trick" of rewriting $2y+3x=xy$ as $(x-2)(y-3)=6$ is nice. Note its resemblance to its far more important cousin, completing the square.

But that's not what this answer is about. We claim that even someone totally innocent of algebra can find all the solutions of $\dfrac{2}{x}+\dfrac{3}{y}=1$, and prove that all of them have been found. All one needs to do is to engage with the numbers.

The key observation is that $x$ and $y$ cannot both be big. We will use this observation fairly efficiently. But even an inefficient approach leads quickly to a full answer.

If $x\ge 5$ and $y\ge 5$, we are in trouble unless $x=y=5$. This satisfies the equation. Except in this case, one of $x$ or $y$ must be $\le 4$. A short list of candidates! And many of them are obviously no good.

Can $x=1$? Of course not, the result would be too big. Similarly, $x$ cannot be $2$.

If $x=3$, then $y=9$. If $x=4$, then $y=6$.

Can $y=1$, $2$, or $3$? Of course not. Put $y=4$. We get $x=8$.

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I initially did that too. But I thought they meant something else with this problem, immediately after I posted I realized that this problem is not much of a problem at all.. – JohnPhteven Feb 14 '13 at 21:28
@ZafarS: As a result, you learned an algebraic trick, so it all turned out well. – André Nicolas Feb 14 '13 at 21:40
In my opinion, this is the best answer. One can easily miss this simple idea and instead attempt to bash it with a standard factorization. – Potato Feb 14 '13 at 23:00
Thank you. But you explained the trick well. – André Nicolas Feb 14 '13 at 23:09

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