In hexadecimal calculus, the square of a positive integer x is 2 identical blocks each of length k. can k be equal to 2023?
I transferred the required number to the decimal number system and got ab ... yzab ... yz (16) = (z * 1 + y * 16 + ... + a16 ^ 2022) * (16 ^ 2023 + 1), (10)I also realized that if the number in the hexadecimal number system is a square, then in the decimal number system it is also a square.
Yes. Assuming you don't allow leading zeros, you want a square of the form $$ x^2 = (16^{2023}+1) b$$ where $16^{2022} \le b < 16^{2023}$. Now it turns out that $16^{2023} + 1$ is divisible by $17^2$ (this is true because $2023$ is odd and divisible by $17$). Thus we can take $$ x = \frac{(16^{2023}+1)\cdot 16}{17}, \ b = \frac{(16^{2023}+1) \cdot 16^2}{17^2} $$