How to write inverse of integer as sum of fractions I was reading this article about partial fractions and at the bottom of the article there was a paragraph about integers. However, I cannot seem to get it right each time.
For example:
1/8 = 1/2^3 = 1/2 - 1/4 - 1/8

but
1/9 = 1/3^2 ≠ 1/3 - 1/9

 A: The summation formula for a finite geometric series is
$$\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}=\frac{1-a^n}{1-a}\tag{1}$$
or
$$\sum_{k=0}^{n-1}a^{-k}=\frac{1-a^{-n}}{1-a^{-1}}=\frac{a^n-1}{(a-1)\,a^{n-1}}$$
In particular, when $a=2$ in the top formula, we can express $2^n-1$ as a sum of lower powers of $2$. But for integers $a>2$ (like $a=3$), the denominator $a-1$ is not $1$.

$$\frac{x}{a}+\frac{y}{b}=\frac{c}{d}\tag{2}$$
is solvable in integers (for $x,y$ given $a,b,c,d$)
iff $d$ divides $mc$ for $m=\href{http://en.wikipedia.org/wiki/Least_common_multiple}{\operatorname{lcm}}(a,b)$, for then the equation becomes
$$\frac{m}{a}x+\frac{m}{b}y=\frac{mc}{d}$$
or, with
$g=\href{http://en.wikipedia.org/wiki/Greatest_common_divisor}{\operatorname{gcd}}(a,b)=\frac{ab}{m}$,
$a'=\frac{m}{a}=\frac{b}{g}$, $b'=\frac{m}{b}=\frac{a}{g}$ and $c'=\frac{mc}{d}=\frac{abc}{dg}$, the famous linear diophantine equation
$$a'x+b'y=c'$$
which is solvable (with infinitely many solutions) since $\operatorname{gcd}(a',b')=\operatorname{gcd}(\frac{b}{g},\frac{a}{g})=\frac{\operatorname{gcd}(a,b)}{g}=\frac{g}{g}=1$.
As we increase the number of terms on the left in $(2)$,
we can of course represent more numbers $\frac{c}{d}$,
but we will always hit a brick wall when a prime $p$
(or prime power $p^k$) dividing $d$ does not also divide
the product of $c$ with the LCM of the denominators on the LHS.
A: $\frac{1}{9} = \frac{a}{3} + \frac{b}{9}$
That implies, $3a + b =1$.
put a and b values and satisfies the above equation.
