<s>[INST] Sum of the reciprocals of several semi-power numbers (such as $\frac{2}{5}=\frac{1}{5}+\frac{1}{5},\frac{1}{30}=\frac{1}{48}+\frac{1}{80}$) If $n$ is a positive integer with at most two different prime factors, then we called $n$ a semi-power number. Hence $2,3,4,6,12,\cdots$ are all semi-power numbers, but $30=2\cdot3\cdot5$ is not.
Is it true that every positive rational number can be expressed as the sum of the reciprocals of several semi-power numbers? (Such as $\frac{2}{5}=\frac{1}{5}+\frac{1}{5},\frac{1}{30}=\frac{1}{48}+\frac{1}{80}$.)
 [/INST] We need a fact from linear Diophantine equations: if $a,b$ are positive and relatively prime and $n>ab$, then there exist positive integers $x,y$ with
$$xa + yb = n.$$
I can sketch a proof if needed.
Now suppose $p$ is prime and does not divide two positive integers $a,b$, and $m$ is a positive integer. Then for some positive integer $n$, $b^n > abp^m$ and so there exist positive integers $x, y$ with 
$$xab + yp^m = b^n.$$
Dividing through gives
$$\frac{x}{p^mb^n} + \frac{y}{ab^{n+1}} = \frac{1}{abp^m}.$$
The above plus induction on the number of prime factors in the denominator gives you your result. It is enough to show that $\frac{1}{n}$ can be written as the sum of reciprocals semi-power numbers.
Base case: $n$ contains one or two prime factors. Then $n$ is semi-power and $\frac{1}{n}$ is the reciprocal of one.
Inductive case: Suppose $\frac{1}{n}$ is the sum of reciprocals of semi-power numbers for all integers $n$ with up to $d$ prime factors. Suppose $n$ has $d+1>2$ prime factors; then $n$ can be written as
$$n = p^m a b$$
where $ab$ contains $d$ prime factors (and $\gcd(p,ab)=1$). The above shows how to write $\frac{1}{n}$ as the sum of fractions whose denominator contains at most $d$ prime factors.
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