Find all triplets (x, y, z) of positive integers such that 1/x + 1/y + 1/z = 4/5

Question

Here's what i did :-

i wrote Find all triplet $$1/x + 1/y + 1/z = 4/5$$ as $$1/x + 1/y = 4/5 - 1/z$$

$(4z-5)/5z = 1/x + 1/y$ which can be written as $1/x + 1/y = 1/(5z)/(4z-5)$

Let $(5z)/(4z-5) = n$ therefore $1/x + 1/y = 1/n$ which on factorization gives $(n-x)(n-y) = n^2$ and $n$ should be an integer , i am not sure about the proof but i got $(5z)/(4z-5)$ is never an integer except $z = <o,1>$ but z cannot be 0 .

Here's my problem :- if we take $n = 1$ we get 0 solutions of $<x,y,-1>$ is what i did correct ?

Answer 1

I am puzzled that you set $\frac{5z}{4z-5}=n$.

Assuming $x \le y \le z$, we have

$$\frac45 \le \frac3x$$

and hence we can conclude that $x \le \frac{15}{4}<4.$ Furthermore, clearly, we need $x > 1$.

Hence $x$ is either $2$ or $3$.

Case $1$:

• If $x=3$,$\frac1y + \frac1z = \frac45-\frac13=\frac7{15}$.

$$15(y+z)=7yz$$

$$7yz-15y-15z=0$$ $$7(7yz-15y-15z)=0$$ $$(7y-15)(7z-15)=225$$

Consider the factors of

\begin{align} 225 &= 1 \times 225 \\ &= 3 \times 75 \\ &= 5 \times 45 \\ &= 9 \times 25 \\ &= 15 \times 15 \end{align}

Since $x=3$, $7y-15 \ge 7x-15 \ge 6$. Also, it is easy to check that $7y-15$ can't take values $9$ or $15$. Hence we conclude that $x \ne 3$.

Hence $x=2$.

Case $2$:

• $x=2$.

$$\frac1y+\frac1z=\frac45-\frac12=\frac3{10}$$

$$3yz-10y-10z=0$$

$$3(3yz-10y-10z)=0$$

$$(3y-10)(3z-10)=100$$

\begin{align} 100 &= 1 \times 100 \\ &= 2 \times 50 \\ &= 5 \times 20\\ &= 10 \times 10 \end{align}

Note that $3y-10\ge -4$ and $3z-10 \ge -4$.

It is easy to see that $3y-10 \in \{2,5\}$.

Hence $y \in \{4,5\}$.

If $y=4$, $z=20$. If $y=5, z=10.$

Now examine $x=2, y=4$, we have $\frac1x+\frac1y=\frac34$. The assumption of it is of the form of $\frac1n$ is not valid.

Answer 2

Hint: Say $x\leq y\leq z$ then ${1\over z}\leq {1\over y}\leq {1\over x}$ so $${3\over x}\geq {4\over 5} \implies x\leq {15\over 4} \implies x\leq 3$$

So do the cases $x=1$ (clearly we can rule out this case), $x=2$ and $x=3$ (for each repeat the procedure I just did on two variables.)

Answer 3

To get a pro bono Diophantine equation, multiply both sides by $5xyz$:

$5yz+5xz+5xy=4xyz$

Reduce (mod 5) to get $4xyz \equiv 0$. Assuming $x\leq y,z$, we have $4/5\leq 3/x$, so $x <4$. Let us take $y$ to be a variable divisible by 5.

Case $x=2$:

$y=5$ implies $z=10$.

$y=10$ implies $z=5$.

Otherwise, $z<5$. Since $1/z\leq 4/5-1/x$, then $z>3$.

$z=4$ implies $y=20$.

Case $x=3$.

We have $\frac 1y+\frac 1z=\frac 45-\frac 1x=\frac 7{15}$.

Multiplying by $15xy$, we get

$15z+15y=7yz$.

Reduce mod 3: $0\equiv yz$ -- so, either $y$ or $z$ divisible by $3$.

Case $y$ divisible by 3: we already assumed $y$ divisible by $5$; so $y$ divisible by 15.

So, $\frac 1z=\frac 7{15} - \frac 1y\geq \frac 7{15}-\frac 1{15}=\frac 25$, or $z\leq 2.5$. Since $z\geq x$, then $z=2$, and clearly this does not work.

Case $z$ divisible by $3$:

As above, $\frac 1z=\frac 7{15} - \frac 1y\geq \frac 7{15}-\frac 15=\frac 4{15}$, so $z\leq 15/4$. Then $z=3$ is the only choice, and again this does not work.

Answer 4

Just to give another way (I like that). It is known the only $n$ such that its decimal expression is finite are $n=2^a\cdot5^b$ Since we look for $$\frac 1x+\frac 1y+\frac 1z=0.8$$ we begin with $x=2$ which gives $$\frac 1y+\frac 1z=\frac45-\frac12=\frac{3}{10}=\frac{6}{20}$$ from which the solutions $$(x,y,z)=(2,5,10),(2,4,20)$$ and obviously all the permutations so the smallest integer to be considered is $x=8$.

It is easily verified that all other such as $x=8,16,25,....$ are not solutions.