## 4 Results  
### Using Algebraic Extensions to Construct Orthogonal Bases  
Recall that an algebraic extension is formed by adjoining, to the ground field, the root of an irreducible equation, i.e., an equation that has no solution in the ground field. In particular, the complex numbers \(x=a_{1}+ia_{2}\) are formed by adjoining to the set of real numbers the root of the equation  
\[x^{2}+1=0 \tag{20}\]  
which has no real roots. Galois fields \(GF(p^{n})\) are constructed by adjoining to the field \(GF(p)\) the root \(\theta\) of an irreducible equation of degree \(n\). Let us consider as an example the construction of the fields \(GF\big{(}\xi^{2}\big{)}\) and \(GF\big{(}7^{2}\big{)}\). The field \(GF(5^{2})\) is constructed as an algebraic extension of the field \(GF(5)\). The elements \(0,1,2,3,4\) are chosen as the elements of the field \(GF(5)\). Equation (20) in the field \(GF(5)\) has a solution. Therefore, the irreducible equation is used [19]  
\[x^{2}=3 \tag{21}\]  
Element 3 in the field \(GF(5)\) is the inverse of element 2. Accordingly, if we select elements of this field in the form \(-2,-1,\ 0,\ 1,\ 2\), then instead of Equation (21), we could use the equation \(x^{2}+2=0\), similar to (20). If there are no negative numbers in the set used, then the correct form is (21). Accordingly, the elements of the field \(GF\big{(}5^{2}\big{)}\) can be represented in the form [57]  
\[x=a_{1}+ia_{2} \tag{22}\]  
where \(i\) is the root of the irreducible Equation (21), which can also be interpreted as a logical imaginary unit. The multiplication of elements of the form (22), therefore, is carried out according to the rule  
\[xy=(a_{1}+ia_{2})(b_{1}+ib_{2})=(a_{1}b_{1}+3a_{2}b_{2})+i(a_{2}b_{1}+a_{1}b_{ 2}) \tag{23}\]  
where the calculation of expressions in brackets is carried out in the sense of the field \(GF(5)\). The field \(GF\big{(}7^{2}\big{)}\) is constructed as an algebraic extension of the field \(GF(7)\). The elements \(0,1,2,3,4,5,6\) are chosen as the elements of the field \(GF(7)\). In this field, Equation (20) is irreducible; therefore, for this case, \(i^{2}=-1\). However, if negative elements are not used, then the correct form is \(x^{2}=6\). Accordingly, the multiplication of the field elements is carried out according to the rule  
\[xy=(a_{1}+ia_{2})(b_{1}+ib_{2})=(a_{1}b_{1}+6a_{2}b_{2})+i(a_{2}b_{1}+a_{1}b_{ 2}) \tag{24}\]  
where the calculation of expressions in brackets is carried out in the sense of the field \(GF(7)\). Orthogonal bases in the fields \(GF\big{(}5^{2}\big{)}\) and \(GF\big{(}7^{2}\big{)}\) can also be constructed according to rule (13). The first elements of such sequences are all nonzero elements of the fields under consideration that satisfy the equation  
\[x^{q-1}=1 \tag{25}\]  
where \(q=p^{n}\) is the number of elements of the field, including zero. Therefore, the length of such sequences for the field \(GF\big{(}5^{2}\big{)}\) is 24 cycles, and for the field \(GF\big{(}7^{2}\big{)}\)--48 cycles. However, there is an important nuance. Sequences of (13) obviously have different periodicity [58]. This can be proven as follows. All elements of any Galois field satisfy Equation (25). The number \(q-1\) is not necessarily prime. In particular, it can contain a factor \(q_{1}\). This means that there are \(\frac{q-1}{q_{1}}\) elements satisfying the equation  
\[x^{\frac{q-1}{q_{1}}}-1=0 \tag{26}\]  
The roots of this equation are obviously representable as \(\theta^{q_{1}}\), where \(\theta\) is a primitive element of the field under consideration. All such quantities appear as the first elements of sequences (13) in their general list. These sequences have a period that contains not \(q-1\), but \(\frac{q-1}{q_{1}}\) cycles. In particular, for the case of the field \(GF\big{(}7^{2}\big{)}\), the value \(q-1=48\). This number contains the factor \(q_{1}=2\), i.e., among sequences of the form (13) in this case there are 24 sequences that have a period containing 24 cycles (or less).