Euge said:Hi Prefer,
Please show what your thoughts are on this problem or where you're stuck. Have you been able to prove one of the directions, e.g., if the equation $ y^2 = r $ has a solution, then $ r^m = 1$?
Prefer said:I have the direction you mention, so far:
The non-zero elements of $\mathbf F_{q^n}$ are a cyclic group of order $q^n-1=2m$. So, if $y^2=r$, we have $r^m=y^{2m}=1$.
The concept of "squares" in a field with q^n elements is related to the properties of the field's elements when operated under multiplication. In this context, a square is any element that can be obtained by multiplying another element by itself.
The number of squares in a field with q^n elements is equal to half of the total number of elements in the field. This is because every element has a unique square, and the other half of the elements are squares themselves.
The number of squares in a field with q^n elements is directly related to the field's order. More specifically, the number of squares is equal to q^(n-1). In other words, the number of squares is one less than the order of the field.
In a field with q^n elements, squares behave differently under addition compared to multiplication. While squares are preserved under multiplication, they are not necessarily preserved under addition. This means that when adding two squares, the result may not be a square.
Squares in a field with q^n elements have many practical applications in fields such as coding theory, cryptography, and error-correcting codes. In these fields, squares play a crucial role in constructing efficient and secure algorithms and protocols.