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The equation:

$${2 \over x} + {3 \over y} = {5 \over z} ,\,\,x < z$$

$x,y,$ and $z$ must be natural numbers.

How would I go about proving this equation possible to solve?

I can't seem to figure out how to prove it without doing some "random" calculations to see how numbers fit in the equation.


The variables for the possibly lowest answer is: $$x=1,y=6,z=2$$

Since: $$ {2 \over 1} + {3 \over 6} = {5 \over 2} ,\,\,1 < 2 $$

How would it be possible to find this answer, or any other correct answer for that matter, without using "random" placeholder variables to see how to equation evolves?

Simpler put: How to is it possible to solve this equation faster and more effectively? Is it possible to determine whether or not the equation is solvable before trying to solve it?

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Essentially you tried $x=1$ and were able to get a solution quickly, so it's hard to argue with success. Knowing we want $x \lt z$, I thought about making both sides equal to 1, i.e. $z=5$ and finding off the top of my head $x=3$ and $y=9$. My values $x,y,z$ are bigger, but the equal sides of the equation are smaller than in your solution. – hardmath Apr 23 '13 at 16:26
up vote 4 down vote accepted

Here's a thought. (Not a very general method, but still useful, I hope.) Rewrite the equation as follows $$\begin{align}\frac3y &= \frac5z - \frac2x\\\frac3y &= \frac{5x-2z}{xz}\\y&=\frac{3xz}{5x-2z}\tag{*}\end{align}$$ Note that in case where $x,z$ are integers such that $5x-2z=1$, we have that $y$ is automatically an integer. But this means that in this case $5x=2z+1$, so $2z+1$ is an odd multiple of $5$. In other words, every solution of $5x=2z+1$ is of the form $$\begin{align}x&=2k+1\\z&=5k+2\end{align}$$ for some integer $k\in\mathbb Z$. Note that $z>x$ iff $k\in\mathbb N_0$. This yields an infinite family of solutions: $$\begin{align}x&=2k+1\\y&=3(2k+1)(5k+2)\\z&=5k+2\end{align}$$ where $k\in\mathbb N_0$ is arbitrary.

In $(*)$ you could also examine $5x-2z=l$ for some other values of $l$, which would yield solutions in case if $3xz$ were divisible by $l$. A particularly nice case to examine would be $5x-2z=3$ in which case, as in the case we examined, this happens automatically. (Other easy cases are $l=-1,-3$, which are the remaining divisors of 3.)

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Firstly if $x<z$ then $y>z$.Because if $y<z$,$$\frac{2}{x}+\frac{3}{y}>\frac{2}{z}+\frac{3}{z}=\frac{5}{z}$$Contradiction!

So let $x=z-a$ and $y=z+b$.Plug these values in the equation and you will get,$$z=\frac{5ab}{3b-2a}=\frac{5}{\frac{3}{a}-\frac{2}{b}}$$

Hence minimum value of z can be 5. Therefore, $$\frac{3}{a}-\frac{2}{b}=1\implies b=\frac{2a}{3-a}=\frac{2}{\frac{3}{a}-1}$$

The minimum value of $\frac{3}{a}-1$ is 2 (because if $\frac{3}{a}-1=1$ a = 1.5(a fraction))

Hence minimum value of b=2/2=1.$$\frac{3}{a}-1=2\implies a=1$$ So,$$x=z-a=5-1=4$$ $$y=z+b=5+1=6$$

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Why do you say the minimum value of $z$ is $5$? – Erick Wong Apr 23 '13 at 23:41
Both the OP answer $(x,y,z)=(1,6,2)$ and the answer $(2,12,4)$ have $z<5$ (just to back up Eric's comment). However these are the only two for which $z<5$. – coffeemath Apr 24 '13 at 1:29

This is an approach which shows a way to (in theory) obtain all the solutions. We start with your equation [*]: $2/x+3/y=5/z$ and, since your restriction is $x<z$, multiply by $z$ to obtain $$2\frac{z}{x}+3\frac{z}{y}=5.$$ Not much of a start, but now note that if we define the rational variables $s=z/x,\ t=z/y$ we have $$[1]\ \ 2s+3t=5,$$ and our requirement that $x<z$ is now the requirement that $s>1$. Then from $2s+3t=5$ and $t>0$ it follows that also $s<5/2$.

This suggests a procedure: Select any rational number $s \in (1,\frac{5}{2})$. Then define $t=(5-2s)/3$, positive given our restriction on $s$, which makes $[1]$ hold, and in fact we'll have $t \in (0,1).$

But how do we obtain integer solutions to [*] from such a rational solution to $[1]$? The idea is to express $s,t$ as fractions $s=a/b,\ t=c/d$ and if it happens that $a \neq c$ we may multiply top and bottom of either or both fractions until they do have equal numerators. This is like putting them over a common denominator, but in reverse, we put them under a common numerator. After this adjustment we now have $s=z/x,\ t=z/y,$ and we have not disturbed the requirement $x<z$ since multiplying top and bottom of $s$ by the same natural number preserves that the numerator exceed the denominator. Tracing things back we arrive at a solution to [*]. All primitive solutions arise this way, and other solutions can be obtained from this by multiplying each of $x,y,z$ by a natural number.

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