Let $A$ be a real skew-symmetric matrix with integer entries. Show that $\operatorname{det}{A}$ is square of an integer.
Here is my idea: If $A$ is skew-symmetric matrix of odd order, then $\operatorname{det}{A}$ is zero. So, take $A$ to be of even order and non-singular. Since all the eigenvalues of $A$ are of the form $ia$ and its conjugate (where $a$ is real number), we see that $\operatorname{det}{A}$ is square of a real number. But I am not getting how to show it is square of an integer.
A proof by induction is given in David J. Buontempo, The determinant of a skew-symmetric matrix, The Mathematical Gazette, Vol. 66, No. 435, Mar., 1982, Note 66.15, pages 67-69. If you have access to jstor, it's here. The proof does not depend on the Pfaffian.
For a skew symmetric $A$, $\det(A)={\rm pfaffian}(A)^2$ where pfaffian is an integral polynomial function of the entries of the matrix $A$. For the case of an integer matrix the pfaffian is therefore an integer. Hence the result you want.
The question is about the determinant of an invertible skew-symmetric $n\times n$ matrix $A$ with integer entries. Work over the field $\mathbb Q$ and write $n=2\cdot r$. First observe that $\det\left(\begin{bmatrix}\mathbf 0 & I_r \\ - I_r & \mathbf 0\end{bmatrix}\right)=\det\left(\begin{bmatrix} I_r&\mathbf 0 \\ \mathbf 0&- I_r \end{bmatrix}\right)\det\left(\begin{bmatrix}\mathbf 0 & I_r \\ I_r & \mathbf 0\end{bmatrix}\right)=(-1)^r(-1)^r=1$
Now write $B^TAB =\begin{bmatrix}\mathbf 0 & I_r \\ - I_r & \mathbf 0\end{bmatrix}$ i.e. $A$ is congruent to the symplectic matrix; this implies $A$'s determinant is positive.
Taking determinants we get $\det\big(A\big) = \det\big(B^{-1}\big)^2$ where $\det\big(A\big)\in \mathbb N$ (as $A$'s components are integers). If $\big\vert \det\big(B^{-1}\big)\big\vert=\lambda$ and $\lambda \not \in \mathbb N$ then $\lambda=\frac{m}{k p}$ where $m$ modulo (prime) $p$ is non-zero (Fundamental Theorem of Arithmetic) which, after clearing denominators and squaring, implies $k^2 p^2 \cdot \det\big(A\big) =m^2$ but the RHS is non-zero mod $p$ (a unit in $\mathbb F_p$) and the LHS is zero mod $p$ which is impossible. Conclude $\lambda \in \mathbb N$ and $\lambda^2=\det\big(A\big)$ is a squared integer.