Let $A=(a_{ij})$ be an $n\times n$ matrix whose each entry is a natural number. Let $B=(b_{ij})$ be obtained from $A$ by setting $$b_{ij}=\begin{cases}0 & \text{if }a_{ij} \text{ is even},\\1 & \text{otherwise}.\end{cases}$$ It is known that both $|A|$ and $|B|$ have same parity. Meaning, determinants of $A$ and $B$ are either simultaneously even or simultaneously odd.
But, I do NOT know a proof to this result. So please suggest a proof or give a reference to the proof.
Note that the determinant of a matrix is simply a polynomial in the entries of the matrix. Remember that the projection $\pi : \mathbb{Z} \to \mathbb{Z}_2$ of a number onto its parity is a ring homomorphism, so $\pi(a + b) = \pi(a) + \pi(b)$ and $\pi(a \cdot b) = \pi(a) \cdot \pi(b)$. From this, it follows that the parity of any polynomial expression of numbers $a_1, \dots, a_n$ depends only on the parity of the numbers, and not on the numbers themselves. So the parity of the determinant of a matrix $A$ depends only on the parity of the entries.
Replacing the matrix $A$ with the matrix $B$ corresponds to replacing every even entry in $A$ with the even number $0$ and every odd entry with the odd number $1$. By the above reasoning, this does not change the parity of the determinant.