Parametric solution of the Diophantine equation $\frac{3}{n}=\sum\frac{1}{a}$

Question

Assmue $n>3$ is a odd number,Prove that there exists distinct odd numbers $a,b,c$ such $$\dfrac{3}{n}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\tag{1}$$ I'm reading a lot about the Erdös-Straus Conjecture (ESC), a conjecture that states that for every natural number p≥2, there exists a set of natural numbers $a,b,c$, such that the following equation is satisfied: $$\dfrac{4}{p}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$ see Egyptian fraction But still I have no idea about how to attack the one $(1)$

Answer 1

This provides only a solution for $n$ not both prime and $-1\mod 12$, as well as how I derived it.

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Let's try to find a couple such triples first. \begin{align} \frac35&=\frac13+\frac15+\frac1{15}\\ \frac37&=\frac13+\frac1{11}+\frac1{231}\\ \frac39&=\frac15+\frac19+\frac1{45} \end{align} We see that in the first and third case, we used a fraction $\frac1n$ so we can try to solve

$$\frac2n=\frac1a+\frac1b$$

instead. The above equation reduces to $n=\frac{2ab}{a+b}$ so we'll solve $a+b\mid 2ab$ for odd $a,b$. First let $\gcd(a,b)=d$ so that $d\mid a+b$. Now note that $a+b\mid 2ab-2b(a+b)$ so $a+b\mid 2b^2$ and so also $a+b\mid 2a^2$. From those two follows:

$$a+b\mid \gcd(2a^2,2b^2)=2\gcd(a^2,b^2)=2\gcd(a,b)^2=2d^2$$

Now let $x=a+b$. We have $d\mid x$ and $x\mid 2d^2$. We write $dk=x$ and $x\gamma=2d^2$ and solving for $x$ and $d$ yields $d=\frac{k\gamma}2$ and $x=\frac{\gamma k^2}2$. We can write $a=\alpha d$ and $b=\beta d$, and using the above expressions we see $k=\alpha+\beta$, which we can substitute into the expression for $d$, after which we substite that expression into $a=d\alpha$ and $b=d\beta$ to get that \begin{align} a=\tfrac12\gamma\alpha(\alpha+\beta)\\ b=\tfrac12\gamma\beta(\alpha+\beta) \end{align} where $\alpha,\beta,\gamma$ are all odd. Now we see that

$$\frac2n=\frac1a+\frac1b=\frac{\alpha+\beta}{\tfrac12\gamma\alpha\beta(\alpha+\beta)}=\frac2{\alpha\beta\gamma}$$

so that we need to write $n=\alpha\beta\gamma$. Note that for $a$ and $b$ to be distinct, $\alpha$ and $\beta$ need to be distinct - and we wanted that, so let's choose $\alpha=1$. Also, since we don't want $b$ to be equal to $n$ (since we actuall chose $c=n$, the variable that we chose at the very start), we need $n$ to be composite, since then we can choose $b$ a divisor of $n$ not equal to $1$ or $n$ itself (and since $n$ should be odd, those divisors will also be odd automatically). Now let $n=rs$, where $r,s$ are both positive integers, and $r\equiv1\mod 4$, but also $r\neq1$, then we have a solution

$$\frac3{rs}=\frac1{rs}+\frac{1}{\tfrac12s(r+1)}+\frac1{\tfrac12rs(r+1)}$$

and since $r\equiv 1\mod4$, $r+1$ is only divisible by $2$ once, so that $\tfrac12(r+1)$ is odd, and so $\tfrac12s(r+1)$ and $\tfrac12rs(r+1)$ are odd, so this is a solution.

So we handled all $n$ that have at least one divisor $1\mod 4$ (that divisor not being equal to $1$) and so we're left with all $n$ that are prime and $3\mod 4$.

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Let's try some examples where $n\equiv3\mod4$ is prime. \begin{align} \frac37=\frac13+\frac1{11}+\frac1{231}\\ \frac3{11}=\frac15+\frac1{15}+\frac1{165}\\ \frac3{19}=\frac17+\frac1{67}+\frac1{8911} \end{align} What we see here is that $\frac37-\frac13=\frac2{21}$, and $\frac3{19}-\frac17=\frac2{133}$. Since we already solved $\frac2x=\frac1a+\frac1b$, we can use this; let $n=12p+7$ be prime. Now we can choose $c=4p+3$ and see \begin{align} \frac{3}{12p+7}-\frac1{4p+3}&=\frac{12p+9}{(12p+7)(4p+3)}-\frac{12p+7}{(12p+7)(4p+3)}\\ &=\frac{2}{(12p+7)(4p+3)} \end{align} Which we know is solvable, since we can just look at our solution above to that kind of equation, and find $a=\tfrac12(12p+7)(4p+3)((12p+7)(4p+3)+1)$ and $b=\tfrac12((12p+7)(4p+3)+1)$.

Now we've narrowed it down to primes $-1\mod12$.

Answer 2

It was necessary to write the solution in a more General form:

$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$t,q$ - integers.

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=2qL$

The solutions have the form:

$$x=\frac{p(p-s)}{tL-q}$$

$$y=\frac{p(p+s)}{tL-q}$$

$$z=L$$

Decomposing on the factors as follows: $p^2-s^2=(p-s)(p+s)=qL$

The solutions have the form:

$$x=\frac{2p(p-s)}{tL-q}$$

$$y=\frac{2p(p+s)}{tL-q}$$

$$z=L$$

To find solutions, you can always write. $3L-q=1 ; 2$

Although formally, this problem reduces to the factorization. To such a formula.

$$\frac{t}{q}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

$$k=\frac{qL(p+s)}{ps(tL-q)}$$

$$x=pk$$

$$y=sk$$

$$z=L$$

The calculation procedure is simple. To try all possible options ranging from $1$.

$$tL-q=1;2;3;4;5....$$

A large bust is not needed. It suces to restrict our if the number obtained is less than $2$.