<s>[INST] Prove the inverse of the Hilbert matrix has integer entries 
$1 \frac{1}{2} ... \frac{1}{n}$
$\frac{1}{2} \frac{1}{3} ... \frac{1}{n+1}$
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$\frac{1}{n} \frac{1}{n+1} ... \frac{1}{2n-1}$
Does the inverse of this matrix have integer entries? Prove your
  statement.

I thought of  multiplying every line $i$ by $i!$, but it leads to a very complicated solution (if it leads to a solution at all).
Source: Linear Algebra, Kenneth Hoffman and Ray Kunze. Section 1.6, exercise 12.
 [/INST] The answer to your first question is yes.
Do not dismiss the example referenced in the problem statement (the case where $n = 3$). Examples like the one I'll reproduce below help to build insight:

Given $\quad A=\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\end{pmatrix},\qquad$
  $A^{-1}=\begin{pmatrix} 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 \end{pmatrix}$.

I'd suggest you explore the matrix and its inverse for $n = 4$.
What we want to show is that
$B=\begin{pmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1} \\ \ldots & \ldots & & \ldots \\ \frac{1}{n} & \frac{1}{n+1} & \ldots & \frac{1}{2n-1}\end{pmatrix}$
is invertible and $B^{-1}$ has integer entries. 
Hints to get you started: The matrix $B$ is known as a Hilbert matrix and the entries of its inverse can be represented as the product of binomial coefficients. 
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