How many pairs $(a,b)$ of non-zero real numbers satisfy the equation
\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}\]
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$
$\text{(E)} \ \text{two pairs for each} ~b \neq 0$
We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$.
\[a+b=\frac{ab}{a+b}\]
\[a^2+2ab+b^2=ab\]
\[a^2+ab+b^2=0.\]
By the quadratic formula and checking the discriminant (imagining one of the variables to be constant), we see that this has no real solutions. Thus the answer is $\boxed{\text{(A)none}}$.