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I have a question regarding Terence Tao article on the number of solutions to $\frac{4}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$. In his second lemma he states that the equation $\frac{4}{p} = \frac{1}{\frac{pc+1}{4}p} + \frac{1}{\frac{pc+1}{4a} \frac{a+b}{c}} + \frac{1}{\frac{pc+1}{4b} \frac{a+b}{c}}$ only has solutions if $p=-c \hbox{ mod } 4ab$ or if $p = -c^{-1} \hbox{ mod } 4ab$, $a$ is coprime to $b$, $4ab$ divides $cp+1$ and $c$ divides $a+b$, taking into account that the denominator has to be natural number. He states no proof for this observation.

I started to only look at the solutions for $p=-c \hbox{ mod } 4ab$, that means we can rewrite $p$ to $4abk-c$, where $k$ is some natural number. Then I rewrote $\frac{pc+1}{4ab}$ to $\frac{(4abk-c)c+1}{4ab}=\frac{4abkc-c^{2}+1}{4ab}=kc-\frac{c^{2}-1}{4ab}$. It is clearly that $c$ has to be a natural number else $\frac{pc+1}{4}$ wouldn't produce a natural number, as $p$ should be a prime. I also proofed that both $a$ and $b$ must be natural numbers.

Now back to $\frac{c^{2}-1}{4ab}$. For this to hold $c$ must be odd, as $4ab$ is even. Thus I wrote $c = 2k+1$ for some integer $k \geq0$. Now we have $\frac{(2k+1)^{2}-1}{4ab}=\frac{4k^{2}+4k}{4ab}=\frac{k^{2}+k}{ab}=\frac{k(k+1)}{ab}$. As $k$ and $k+1$ are always relative prime (you can conclude that from Euclid's theorem), $a$ has to be relative prime to $b$ (if both are square free, see comments please).

Now that's where I'm stuck so far. I can't seem to proof why $c$ should divide $a+b$.

Could someone give me a hint please? Note that I'm still in high school, but I taught myself more advanced mathematical subjects.

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  • $\begingroup$ Note that $ab|k(k+1)$ does not mean that $a$ and $b$ must be relatively prime. Take $k=8$ then $6 \times 12 = 8 \times 9$ $\endgroup$ – Mark Bennet Sep 10 '13 at 12:04
  • $\begingroup$ Oops, you are right, I forgot to add the restriction that $a$ and $b$ must be square free. $\endgroup$ – Jori Sep 10 '13 at 16:25
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[Not really an answer, but too complicated for a hint]

What Tao writes is that if $${4\over p}={1\over n_1}+{1\over n_2}+{1\over n_3}$$ has a solution with exactly one of the $n_i$ divisible by $p$, then that solution is of the form $${4\over p}={1\over{pc+1\over4}p}+{1\over{pc+1\over4a}{a+b\over c}}+{1\over{pc+1\over4b}{a+b\over c}}$$ He notes that similar criteria have been used in a paper of Vaughan that he cites, so maybe there is an explanation to be found in Vaughan's paper.

As to how you might prove it from scratch, we are assuming $n_1$ (say) is a multiple of $p$, say, $n_1=pd$ for some $d$, so we have $${4d-1\over pd}={1\over n_2}+{1\over n_3}$$ Now there is a lot known about how to write a fraction $x/y$ in the form $(1/r)+(1/s)$, and I would try to apply that knowledge to this equation.

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  • $\begingroup$ Here are two possibilities. One is to work out for yourself the details of expressing a rational as a sum of two unit fractions. You may find it a challenging but doable exercise. The other is to go to a library and see whether they will get Vaughan's paper for you on Interlibrary Loan. Or, if there is a university near you, they may already have the publication on their shelves. $\endgroup$ – Gerry Myerson Sep 18 '13 at 5:38

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