The question posed by Marcel Țena on Facebook sparked heated debates:
"How do you write 2023 as the sum of four perfect squares ?"
Without finding an answer, I commented with a challenge. "Petre Ciobanu
Challenge: I learned, I don't know where, that two of the four numbers can always be chosen equally. Who? find this writing ??"
However, the comment did not have any reaction, and I also replied, later: 
So $2023=17^2+23^2+23^2+26^2$. I also mentioned Preda Mihailescu in the comment, who I think is the one who demonstrated the property that in Lagrange's Four Squares Theorem, two numbers can always be chosen equal.
Mr. Vasile MASGRAS published the following list
in total 42 representations.
Among the 109 representations, some are identical. I counted 64, finding some identical ones (marked with black). Then, checking with the correct list of the 61 representations (put by Eugen Paltanea see also link), I found three more that are repeated (marked in green). The leftmost number (written in green) is the number corresponding to the correct table with the 61 representations.
The correct answer is therefore 61 representations of the number 2023 as the sum of four perfect squares.
Of them, 42 variants contain all four distinct numbers. Then 18 variants contain two equal numbers and one has a number repeated three times (marked with red color)
In agreement with Jacobi's Theorem, the number of ways to represent n as the sum of four squares is eight times the sum of the
divisors of
n if
n is odd. Here
$$8*(1+7+17+119+289+2023)=8*2456=19\;648.$$
Let's count them.
A writing with 4 different terms, e.g. $2023=1^2+2^2+13^2+43^2$, is counted, taking into account all $\pm$ possible signs
$$2023=(\pm 1)^2+(\pm 2)^2+(\pm 13)^2+(\pm 43)^2,$$
$2*2*2*2=16$ times, each writing is counted 24 times (the number of arranging the 4 terms, e.g. 2023=43^2+1^2+2^2+13^2). In total $24*16=384$. Most are of this type, $42*384=16\;128.$
Writings with two equal terms, e.g. $2023=1^2+10^2+31^2+31^2$, count 16 times if we take into account the $\pm$ signs, and 12 times each if we take into account all possible arrangements. There are 18 of these in total. So we have in total $18*12*16=3\;456$ writings.
Finally, the only writing with three equal terms, $2023=17^2+17^2+17^2+34^2$ is counted 16 times because of the signs $\pm$, and 4 times because of the permutations of the terms. In total $16*4=64.$
In conclusion, the total number of writings will be $$16128+3456+64=19648.$$
