Proving a necessary condition for an egyptian fraction with odd denominators

Question

In the question posted in Proving an equality involving cyclic sums, I realized that all the possible solutions to the following egyptian fraction with denominators from a set of odd positive integers $S=\{x_1,x_2,x_3,x_4\}$ $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$ have the following property:

Property

There is one $x_j\in S$ such that for every $x_i\in S$, $$\gcd(x_j,x_i)=x_i$$

I copy from the question mentioned the possible solutions of the above egyptian fraction, which have motivated my present question, for you to check:

$$[3, 5, 9, 45] [3, 5, 15, 15] [3, 7, 7, 21] [3, 9, 9, 9] [3, 5, 5, 15] [5, 5, 5, 5]$$

However, I am stuck when looking for a proof of this property as a necessary condition for every solution, other than the fact that the solutions listed are all the possible solutions. I am able to prove that every $x_k$ divides $\prod _{i\neq k} x_i$, but nothing more. Any help on this proof would be welcomed!

Answer 1

Too long to comment.

The solutions listed are not all the possible solutions.

In the following, I'm going to prove that all the possible solutions satisfying $x_2\le x_3\le x_4$ are

$$(x_1,x_2,x_3,x_4)=\color{red}{(9,3,3,9)},\color{red}{(15,3,3,5)},\color{red}{(7,3,3,21)},(3, 5, 9, 45),$$ $$(3, 5, 15, 15),(3, 7, 7, 21),(3, 9, 9, 9),(5, 3, 5, 15),(5, 5, 5, 5).$$

(The solutions in red are not listed in the question.)

Proof :

We have $$3\le x_1,\quad (x_1,x_2)\not=(3,3),\quad 3\le x_2\le x_3\le x_4$$ and $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le\frac{5}{\min(x_1,x_2)}\implies \min(x_1,x_2)\le 5$$

Now, let us separate it into three cases :

Case 1 : $x_1=x_2=5$

Suppose that $x_3\ge 7$. Then, we have $$1=\frac 25+\frac{1}{5}+\frac{1}{x_3}+\frac{1}{x_4}\le\frac 25+\frac{1}{5}+\frac{1}{7}+\frac 17=\frac{31}{35}$$ which is impossible. So, we get $x_3=5$ and $x_4=5$.

Case 2 : $x_2\lt x_1$

Suppose that $x_2=5$. Then, we have $$1 = \frac{2}{x_1}+\frac{1}{5}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{7}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{31}{35}$$which is impossible. So, we get $x_2=3$.

Suppose that $x_1\ge 7$ and $x_3\ge 7$. Then, we have $$1 = \frac{2}{x_1}+\frac 13+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{7}+\frac 13+\frac{1}{7}+\frac{1}{7}=\frac{19}{21}$$ which is impossible. So, we get either $x_1=5$ or $3\le x_3\le 5$.

• If $x_1=5$, then we have$$1=\frac 25+\frac 13+\frac{1}{x_3}+\frac{1}{x_4}\implies \frac{4}{15}=\frac{1}{x_3}+\frac{1}{x_4}$$Multiplying the both sides by $4\times 15\times x_3x_4$ gives$$16x_3x_4-60x_3-60x_4=0$$Adding $15^2$ to the both sides gives $$ (4x_3-15)(4x_4-15)=225$$Since $4x_3-15\equiv 4x_4-15\equiv 5\pmod 8$ and $-3\le 4x_3-15\le 4x_4-15$, we get$$(4x_3-15,4x_4-15)=(5,45)\implies (x_3,x_4)=(5,15)$$

• If $x_3=3$, then we have$$1 = \frac{2}{x_1}+\frac 13+\frac{1}{3}+\frac{1}{x_4}\implies (x_1-6)(x_4-3)=18$$Since $x_1-6$ is odd with $x_1-6\ge -1$ and $x_4-3\ge 2$, we have $$(x_1-6,x_4-3)=(1,18),(3,6),(9,2)\implies (x_1,x_4)=(7,21),(9,9),(15,5)$$

• If $x_3=5$, then we have$$1 = \frac{2}{x_1}+\frac 13+\frac{1}{5}+\frac{1}{x_4}\implies (7x_1-30)(7x_4-15)=450$$Since $7x_1-30\equiv 5\pmod{14}, 7x_4-15\equiv 6\pmod{14}, 7x_1-30\ge 5$ and $7x_4-15\ge 20$, we get $$(7x_1-30,7x_4-15)=(5,90) \implies (x_1,x_4)=(5,15)$$

Case 3 : $x_1\lt x_2$

Suppose that $x_1=5$. Then, we have $$1 = \frac{2}{5}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{5}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{29}{35}$$which is impossible. So, we get $x_1=3$.

Suppose that $x_2\ge 11$. Then, we have $$1 = \frac{2}{3}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\le \frac{2}{3}+\frac{1}{11}+\frac{1}{11}+\frac{1}{11}=\frac{31}{33}$$which is impossible. So, we get $x_2\le 9$.

• If $x_2=5$, then we have$$1=\frac 23+\frac 15+\frac{1}{x_3}+\frac{1}{x_4}\implies (2x_3-15)(2x_4-15)=225$$Since $2x_3-15\equiv 2x_4-15\equiv 3\pmod 4$ and $-5\le 2x_3-15\le 2x_4-15$, we get $$(2x_3-15,2x_4-15)=(3,75),(15,15)\implies (x_3,x_4)=(9,45),(15,15)$$

• If $x_2=7$, then we have$$1=\frac 23+\frac 17+\frac{1}{x_3}+\frac{1}{x_4}\implies (4x_3-21)(4x_4-21)=441$$Since $4x_3-21\equiv 4x_4-21\equiv 7\pmod 8$ and $7\le 4x_3-21\le 4x_4-21$, we get $$(4x_3-21,4x_4-21)=(7,63)\implies (x_3,x_4)=(7,21)$$

• If $x_2=9$, then we have $$1=\frac 23+\frac 19+\frac{1}{x_3}+\frac{1}{x_4}\implies (2x_3-9)(2x_4-9)=81$$Since $2x_3-9\equiv 2x_4-9\equiv 1\pmod 4$ and $9\le 2x_3-9\le 2x_4-9$, we get $$(2x_3-9,2x_4-9)=(9,9)\implies (x_3,x_4)=(9,9).\quad\blacksquare$$

Answer 2

I think I have a proof of the property for most cases, I post it for your revision and looking for ideas for the case left.

Proof

We have that $$1 = \frac{2}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}$$

This is only possible if $1<\min\{x_1,x_2,x_3,x_4\}\leq5$. Therefore, the only two possibilities left are that $\min\{x_1,x_2,x_3,x_4\}=3$ or $\min\{x_1,x_2,x_3,x_4\}=5$. An interesting fact here is that both are prime numbers.

If $\min\{x_1,x_2,x_3,x_4\}=5$, then necessarily the property holds, as necessarily $x_1=x_2=x_3=x_4$ is the only possible solution.

If $\min\{x_1,x_2,x_3,x_4\}=3$, all the inmmediate greater odd numbers are 5, 7, and 9. If $x_1=3$, then at least one of the others denominators must be equal or lesser than 9, because $1 = \frac{2}{3}+\frac{3}{9}$ (in which case the property holds, as all the denominators are different combinations of 3 and 9). Therefore, any other solution other than the ones mentioned implies that some other denominator is either 5 or 7, which are prime numbers.

As I commented when posting the question, it is easy to show that every $x_k$ divides $\prod _{i\neq k} x_i$. For the proof, the numerators of the fractions are irrelevant, so we will assume without loss of generality that $\min\{x_1,x_2,x_3,x_4\}=x_1$. Therefore, $x_1\mid x_2x_3x_4$, where $\mid$ means "divide".

As $x_1$ is prime, then some other denominator must be equal to $x_1$ or some multiple of $x_1$; in any of those cases, the property holds for both denominators. As the numerators of the fractions are irrelevant, we will assume without loss of generality that $x_1\mid x_2$.

Let us assume, also without loss of generality, that $x_3$ is prime (as stated, either 5 or 7). We have that $x_3\mid x_1x_2x_4$. It could happen:

(i)That $x_3\mid x_2$, in which case the property will continue holding for $x_1$, $x_2$ and $x_3$,

(ii) That $x_3\mid x_4$. If this were the case, when considering that $x_4\mid x_1x_2x_3$, we would have the possibility of $x_4\mid x_1x_3$, in which case the property would hold for $x_1$, $x_3$ and $x_4$, and otherwise that $x_4\mid x_3$, which would imply that $x_3=x_4$.

In all the cases exposed where the property holds for three of the four denominators, we can show that necessarily it must hold for all of them. For instance, let us assume that $\gcd(x_1,x_2)=x_1$ and $\gcd(x_3,x_2)=x_3$. Then, multiplying by $x_2$, we would have that $$x_2 = \frac{2x_2}{x_1}+\frac{x_2}{x_2}+\frac{x_2}{x_3}+\frac{x_2}{x_4}$$ As all the fractions are integer numbers, then necessarily we get that $x_4\mid x_2$, and otherwise the equation would not be possible. The same process can be applied for all the mentioned cases where the property holds for three of the four denominators.

The only case left for consideration is the one where we had that $x_1\mid x_2$ and $x_3=x_4$. Substituting $x_4$ for $x_3$, we have this two possible cases, depending on the numerator being in a fraction with $x_1$ or $x_2$ as denominator, or $x_3$ as denominator:

(i) $$x_2 = \frac{2x_2}{x_1}+\frac{x_2}{x_2}+\frac{x_2}{x_3}+\frac{x_2}{x_3}$$ In this case, the equation has solution only if $\frac{2x_2}{x_3}$ is integer, which can only be possible if $x_3\mid x_2$, and in which case the property would hold;

(ii) $$x_2 = \frac{2x_2}{x_3}+\frac{x_2}{x_2}+\frac{x_2}{x_1}+\frac{x_2}{x_3}$$ In this case, the equation has solution only if $\frac{3x_2}{x_3}$ is integer, which can only be possible if $x_3\mid x_2$, and in which case the property would hold.

However, it is still not proved the case $x_4\mid x_2x_3$ when $x_4>x_3$, and $x_2\mid x_1x_4$, when $x_2>x_1$.