XOR of rows and columns of 4x4 magic squares An article by John Conway, Simon Norton, and Alex Ryba, “Frenicle’s 880 Magic Squares,” in Jennifer Beineke and Jason Rosenhouse, eds., The Mathematics of Various Entertaining Subjects, Vol. 2, 2017, chapter 5, reports the "Nimm0" property for 4x4 magic squares with numbers from 0 to 15, with sum constraint on the diagonals too:
the bitwise exclusive OR (XOR) of all the numbers in a row or column is always 0 (see also an example here). This can be verified with an exhaustive test of all 880 squares.
At note 1 they say that a conceptual proof has been found by Friedrich Fitting and Oliviero Giordano Cassani independently.
However I am not able to find the proof on the web.
Somebody can share it here?
Or is able to rediscover it?
 A: Let $R_i$ denote the sum of the numbers in row $i$, $i=1,2,3,4$. Then $R_1+R_2+R_3+R_4 = 0+\dots+15 = \frac{15\cdot 16}{2} = 8\cdot 15$. So each $R_i = 2\cdot 15 \equiv 0 \pmod{2}$. Therefore, the last digit of the XOR sum of any row is $0$. Similarly with the columns.
Now we show that the 2nd to last digit of the XOR sum of any row or column is $0$. For $a \in \mathbb{Z}$, let $\overline{a}$ denote $a \pmod{2} \in \{0,1\}$ and $\tilde{a} = \frac{a-\overline{a}}{2}$. Let $\tilde{R_i}$ denote the sum of $\tilde{a}$ over the $a$ in row $i$. Then $\tilde{R_1}+\tilde{R_2}+\tilde{R_3}+\tilde{R_4} = \frac{1}{2}\frac{15\cdot 16}{2}-\frac{1}{2}\sum_{j=0}^{15} \overline{j} = 4\cdot 15 - 4$. Therefore, assuming each $\tilde{R_i}$ is the same (see below), we get that each $\tilde{R_i} = 14 \equiv 0 \pmod{2}$.
Similarly, if we let $\overline{\overline{a}}$ denote $a\pmod{4} \in \{0,1,2,3\}$, $\tilde{\tilde{a}} = \frac{a-\overline{\overline{a}}}{4}$, and $\tilde{\tilde{R_i}} = \sum_{a \in R_i} \tilde{\tilde{a}}$, then $\sum_i \tilde{\tilde{R_i}} = \frac{1}{4}\frac{15\cdot 16}{2}-\frac{1}{4}24 = 24$, so assuming each $\tilde{\tilde{R_i}}$ is the same (see below), each $\tilde{\tilde{R_i}} = 6$, which is, once again, $0 \pmod{2}$.
Finally, making the $\pmod{8}$ analogues of the definitions made above, we get $\sum_i \tilde{\tilde{\tilde{R_i}}} = \frac{1}{8}\frac{15\cdot 16}{2}-\frac{1}{8}56 = 8$, so assuming each $\tilde{\tilde{\tilde{R_i}}}$ is the same (see below), we get that each $\tilde{\tilde{\tilde{R_i}}} = 2$, which is $0 \pmod{2}$.
We have thus shown that the XOR sum of the numbers in any row or column is $0$. 

Each $\tilde{R_i}$ is equal if and only if $\sum_{a \in R_i} \frac{a-\overline{a}}{2} = \frac{1}{2}\sum_{a \in R_i} a - \frac{1}{2}\sum_{a \in R_i} \overline{a}$ is the same for each $i$, but since the $\sum_{a \in R_i} a$'s are equal, we just need the $\sum_{a \in R_i} \overline{a}$'s to be equal. In other words, we need the mod 2 of the magic square to be a magic square mod 2. It is well known that 4x4 magic squares indeed have this property.
Similarly for $\tilde{\tilde{R_i}}$ and $\tilde{\tilde{\tilde{R_i}}}$, an equivalent condition for them to be independent of $i$ is that the mod 4 (resp. 8) of the magic square is a magic square mod 4 (resp. 8). This is also well-known.
A: The Conway et al. paper gives a complete bibliographical reference for the Fitting paper. I'm sure you could get it from a library, through interlibrary loan. Cassani & Conway published an article, Neumbering, The Mathematical Intelligencer 40(1) · July 2017 so maybe there's something in there. Cassani appears to be at Oxford, you could try to contact him. 
