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Determinant of a real skew-symmetric matrix is square of an integer

I know that in general, a skew-symmetric matrix with indeterminate elements has a determinant that can be written as a square of some multivariable polynomial. How to prove this?

And if I do not know anything about the Pfaffian, can I prove the statement that a skew-symmetric matrix with integer entries has a determinant that is a square of some integer? I mean if someone knows how to prove it without using the Pfaffian?

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Li Xinghe
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This is for the second part, a skew-symmetric matrix with integer entries

First, if $n$ is odd, then since $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$, so $\det(A)= 0$, which is the square of an integer.

Now for $n$ even, we proceed by induction, and will show the statement is true over the rationals. Base case $n=2$ is obvious.

Induction step. Suppose we want to show it for some $n=2k+2$. Consider the entries. We already know that the diagonals satisfy $a_{i, i} = 0$. If all the other entries are also 0, then we are done. Otherwise, WLOG we have $a_{1,2}\neq 0$. We will proceed to calculate the determinant of $A$.

Using only row operations, where we add a rational linear multiple of the second row to the others, we can make the first column to be $(0, a_{2, 1}, 0, 0, \ldots, 0)^T$. Specifically, for row $k\neq 1, 2$, we have $b_{k, i} = a_{k, i} - \dfrac {a_{k, 1}}{a_{2, 1}} a_{2,i}$. Now, we use column operations, where we add a rational linear multiple of the second column to the others, and we can make the first row to be $(0, a_{1, 2}, 0, 0, \ldots, 0)$. Specifically, for column $k\neq 1, 2$, we have $c_{i, k} = b_{i, k} - \dfrac {b_{1, k}}{b_{1, 2}} b_{i, 2}$. Then, $\det(A) = a_{1, 2}^2 \det(C)$, and so it remains to show that $C$ is still a skew symmetric matrix.

For rows and columns 1 and 2, nothing changed throughout, (so those entries are skew symmetric).
For $i=k$, $b_{i,i} = 0- \dfrac {a_{i,1}}{a_{2,1}}a_{2,i}$, $c_{i,i} = b_{i,i} - \dfrac {b_{1,i}}{b_{1, 2}} b_{i, 2} = (0 - \dfrac {a_{i,1}}{a_{2,1}}a_{2,i})- \dfrac {a_{1,i}}{a_{2,i}} a_{i,2}=0$

For $i\neq k$, $c_{i,k} = b_{i, k} - \dfrac {b_{1, k}}{b_{1, 2}} b_{i, 2} = b_{i, k} - \dfrac {a_{1, k}}{a_{1, 2}} a_{i, 2}= (a_{i, k} - \dfrac {a_{i, 1}}{a_{2, 1}} a_{2,k})- \dfrac {a_{1, k}}{a_{1, 2}} a_{i, 2}$. Hence, it follows that $c_{i,k}=-c_{k,i}$.

Hence, $C$ is a skew symmetric matrix.

Now, $A$ has an integer determinant, which is also the square of a rational number, hence is a perfect square.

Calvin Lin
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    @adamW I came up with this looking at the 4 by 4 case. I didn't feel that identifying 0's would help, since it doesn't really affect the general identity. In fact, there was no issue go dividing by 0, since we only divided by $a_{1, 2}$ and $a_{2, 1}$ which we can choose to be non-zero. Furthermore, with the definition of $B$, the 0 entries on the diagonal will have disappeared. – Calvin Lin Jan 13 '13 at 18:37
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    You may have misunderstood, I only meant that I had to realize the zeros were there to understand your statement "expanding along that row will give us the result via the induction hypothesis" – adam W Jan 13 '13 at 18:53
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    @adamW The statement was wrongly written. It should be, if there is $a_{I, J} \neq 0$, then we proceed as below. Notice that in the subsequent paragraphs, we only divided out by $a_{1,2}$. Otherwise, if all $a_{I, J} = 0$ the determinant is 0. – Calvin Lin Jan 13 '13 at 19:28
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    Ah, thank you. I believe it would be perfectly fine to do the inline change, as it is much better in my opinion. I see such edits here happen all the time. – adam W Jan 13 '13 at 19:41
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    My opinion right now is this: our comments regarding the edit should be removed and you can go ahead and remove the Inline edit and end edit from your post. They all are not necessary any longer. – adam W Jan 13 '13 at 19:53
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    +1 very nice. Just a comment: the entries of $C$ will not be integers but this is not a problem. If $r$ is a common denominator then $r^n$ is a square and $r^nC$ has integer entries. – P.. Jan 13 '13 at 20:25
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    @Pambos I stated that the induction was done over the rational matrices. And so the conclusion is that the determinant is the perfect square of a rational. Then, any integer that is the perfect square of a rational number, is also the perfect square of an integer, which is easy to show with the integer root theorem. It's not immediately clear that we can multiply throughout by the common denominator, to achieve the same determinant. – Calvin Lin Jan 13 '13 at 22:41