From Diofantian Equations To Matricial Equations (Ii) -Generalizations Of The Pythagorean Equation-

Generalizations of the Pythagorean equation

Immediate generalization of the Pythagorean equation (A) is given by the equation:

x²+y²+z²=t². (B)

The positive solutions of equation (B) represent the dimensions and the length of the diagonals of a rectangular parallelepiped. (Andreescu & Andrica, 2002) As in the case of the right triangle, here too we are interested in the situation in which all these dimensions are natural numbers.

For a start, three observations are required here as well:

If the quadruple (x₀,y₀,z₀,t₀) satisfies equation (B), then any quadruple of the form (kx₀,ky₀,kz₀,kt₀), with k, is a solution of this in. Therefore, to solve equation (B) it is sufficient to determine the solutions (x,y,z,t) with the property that:

. (5.1.1)

That is why the following definition is required (and here):

A solution (x₀,y₀,z₀,t₀) of equation (B), which verifies the equality (5.1.1) is called a primitive solution.

Quadruples (0,y,z,t), (x,0,z,t) and (x,y,0,t), respectively, with x, y, z and t natural numbers, which reduce equation (B) to an equation of type (A) will not be analyzed in this paper; they can be seen in (Vălcan, 2019).

Therefore, in the following, we will consider only solutions with non-zero components.

Equation (B) is symmetric in x, y and z; so, if (x,y,z,t) is its solution, then also (x,z,y,t), (y,x,z,t), (y,z,x,t), (z,x,y,t), (z,y,x,t) are (also) solutions of the equation (B).

We present below the form of a primitive solution of equation (B):

All primitive solutions in natural numbers of equation (B) are given by the equalities:

x=2a, y=2b, $\text{z}=\frac{a^2+b^2-c^2}{c}$ and $\text{t}=\frac{a^2+b^2+c^2}{c}$, (5.1.2)

where a, b, and c is a divisor of a²+b², which is less than $\sqrt{a^2+b^2}$, and:

(5.1.1)

Any solution is obtained only once in this way.

Let (x,y,z,t) be a solution of equation (B). Based on Remark 2.4 from (Vălcan, 2019), we distinguish the following cases:

All numbers x, y and z are odd. Then t will be odd; so:

x=2a+1,y=2b+1,z=2c+1and t=2d+1, (5.1.3)

with a, b, c, d. From equalities (B) and (5.1.3) it follows that:

4a²+4a+1+4b²+4b+1+4c²+4c+1=4d²+4d+1, (5.1.4)

that is:

4a²+4a+4b²+4b+4c²+4c+2=4d²+4d, (5.1.4)

which is impossible modulo 4; or otherwise: by dividing by 2, the left limb is an odd number and the right limb is an even number.

One of the numbers x, y, z is even; say the number x is even, and the numbers y and z are odd. Then the number t is even:

x=2a,y=2b+1,z=2c+1and t=2d, (5.1.5)

with a, b, c, d. From equalities (B) and (5.1.5) it follows that:

4a²+4b²+4b+1+4c²+4c+1=4d², (5.1.6)

that is:

4a²+4b²+4b+4c²+4c+2=4d², (5.1.6)

which is impossible modulo 4; or otherwise: by dividing the equality (5.1.6) by 2, the left limb is an odd number and the right limb is an even number.

Two of the numbers x, y, z are even; say the numbers x and y are even and the number z is odd. Then the number t is odd, too. So:

x=2a,andy=2b, (5.1.7)

with a, b. From equalities (B) and (5.1.7) it follows that:

4a²+4b²+z²=t². (5.1.8)

We note:

t-z=d. (5.1.9)

Then, from equalities (B), (5.1.8) and (5.1.9), we obtain that:

4a²+4b²+z²=(z+d)², (5.1.8)

whence it follows that:

4a²+4b²-d²=2zd. (5.1.8)

Equality (5.1.8) shows us that d is even, so:

d=2c, (5.1.9)

with d. Now, from equalities (5.1.8) and (5.1.9) it follows that:

$\text{z}=\frac{a^2+b^2-c^2}{c}$. (5.1.10)

Finally, from equalities (5.1.9) and (5.1.10), it follows that:

$\text{t}=\frac{a^2+b^2+c^2}{c}$. (5.1.11)

Because z, t, from any of the equalities (5.1.10) or (5.1.11), it follows that c is a divisor of the number a²+b², and from the equality (5.1.10) it follows that c $\sqrt{a^2+b^2}$.

All numbers x, y and z are even. Then the number t is even and, thus, we contradict the equality (5.1.1), which we considered as a hypothesis.

Numbered t is even. Then one of the numbers x, y, z is even and we reach Case 2 again.

Therefore, any solution is of the form (5.1.2). Of course, c does not divide by b, because otherwise equality (5.1.1) does not exist.

We observe that any solution of the form (5.1.2) of equation (B), with x and y even numbers it is obtained exactly once by the above formulas. Indeed, from what has been proved above, we obtain that:

a = $\frac{x}{2}$,b = $\frac{y}{2}$andc = $\frac{t-z}{2}$;

so the integers a, b and c are uniquely determined by the quadruple (x, y, z, t).

Reciprocally, the following identity:

${\left(2\bullet a\right)}^2+{\left(2\bullet b\right)}^2{\left(\frac{a^2+b^2-c^2}{c}\right)}^2={\left(\frac{a^2+b^2+c^2}{c}\right)}^2$ (5.1.12)

shows that the quadruple defined in the above theorem statement is a solution of equation (B) and, in addition, the numbers x, y are even.

The above theorem not only states the existence of solutions of the given equation, but also provides a practical method for determining all these solutions. It is observed that in order to obtain the solutions of the equation, abstracting from its symmetry in x, y, z it is sufficient to consider only the pairs (a, b), with ab and to take only those c for which z is odd. Thus we eliminate the solutions for which x, y, z, t are all even numbers (Cucurezeanu, 2005).

Two remarks are required here:

The integer solutions of equation (B) can be expressed in the form:

x=2ac,y=2bc,z=a²+b²-c²andt=a²+b²+c², (5.1.13)

where a, b, c.

We notice that in this form - (5.1.13) - it is possible to obtain, for equation (B), this solution several times. On the other hand, this writing has the advantage that it is very similar to that given for the solutions of equation (A) and is easier to remember.

If (x,y,z,t), is the solution of equation (B), then:

a=y+z-t,b=x+z-t,c=x+y-t andd=x+y+z-2t, ()

it is also a solution of the same equation.

A simple calculation shows us that:

a²=y²+z²+t²+2yz-2yt-2zt,

b²=x²+z²+t²+2xz-2xt-2zt,

c²=x²+y²+t²+2xy-2xt-2yt,

d²=x²+y²+z²+4t²+2xy+2xz-4xt+2yz-4yt-4zt,

and:

a²+b²+c²=2x²+2y²+2z²+3t²+2xy+2xz+2yz-4xt-4yt-4zt

=2x²+2y²+2z²+3t²+2xy+2xz+2yz-4xt-4yt-4zt

=x²+y²+z²+4t²+2xy+2xz+2yz-4xt-4yt-4zt+x²+y²+z²-t²,

=d².

Let us now consider equation (B) in the ring (M_n(),+,):

X²+Y²+Z²=T². (B)

The equalities (5.1.13) entitle us to present the following result:

Equation (B) is solvable in the ring (M_n(),+,).

Indeed, we notice that if A, B, CM_n() and verifies the equalities:

AB=BA,BC=CB,AC=CA, (5.1.14)

then the matrices:

X=2AC,Y=2BC,Z=A²+B²-C² andT=A²+B²+C², (5.1.13)

verifies equality (B).

Starting from Example 3.10.1 of (Vălcan, 2019) we can consider the matrices:

A= $\left(\begin{array}{cc}4& 1\\ 3& 4\end{array}\right)$,B= $\left(\begin{array}{cc}2& 5\\ 5& 2\end{array}\right)$ andC= $\left(\begin{array}{cc}5& 2\\ 6& 5\end{array}\right)$.

Then:

A²= $\left(\begin{array}{cc}19& 8\\ 24& 19\end{array}\right)$, B²= $\left(\begin{array}{cc}79& 20\\ 60& 79\end{array}\right)$andC²= $\left(\begin{array}{cc}37& 20\\ 60& 37\end{array}\right)$,

AB= $\left(\begin{array}{cc}23& 22\\ 66& 23\end{array}\right)$=BA,AC= $\left(\begin{array}{cc}26& 13\\ 39& 26\end{array}\right)$=CA andBC= $\left(\begin{array}{cc}40& 29\\ 87& 40\end{array}\right)$=CB;

Now, according to the equalities (5.1.13) we have:

X= $\left(\begin{array}{c}52\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}26\\ 78\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}52\end{array}\right)$,Y= $\left(\begin{array}{cc}80& 58\\ 174& 80\end{array}\right)$,Z= $\left(\begin{array}{cc}61& 8\\ 24& 61\end{array}\right)$,T= $\left(\begin{array}{cc}135& 48\\ 144& 135\end{array}\right)$,

X²= $\left(\begin{array}{c}4732\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2704\\ 8112\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4732\end{array}\right)$,Y²= $\left(\begin{array}{cc}16492& 9280\\ 27840& 16492\end{array}\right)$,

Z²= $\left(\begin{array}{cc}3913& 976\\ 2928& 3913\end{array}\right)$, T²= $\left(\begin{array}{c}25137\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}12960\\ 38880\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}25137\end{array}\right)$,

and the equality (B) is immediately verified.

Let us now consider another example of the solution of the equation (B), of the form (5.1.13), with the matrices A, B, CM₂(); thus be:

A= $\left(\begin{array}{cc}4& 5\\ -3& 1\end{array}\right)$,B= $\left(\begin{array}{cc}3& 5\\ -3& 0\end{array}\right)$ andC= $\left(\begin{array}{cc}2& 5\\ -3& -1\end{array}\right)$.

Then:

A²= $\left(\begin{array}{cc}1& 25\\ -15& -14\end{array}\right)$, B²= $\left(\begin{array}{cc}-6& 15\\ -9& -15\end{array}\right)$andC²= $\left(\begin{array}{cc}-11& 5\\ -3& -14\end{array}\right)$,

AB= $\left(\begin{array}{cc}-3& 20\\ -12& -15\end{array}\right)$=BA,AC= $\left(\begin{array}{cc}-7& 15\\ -9& -16\end{array}\right)$=CA andBC= $\left(\begin{array}{cc}-9& 10\\ -6& -15\end{array}\right)$=CB.

Now, according to the equalities (5.1.13), we have:

$\begin{array}{c}\mathrm{X}=\left(\begin{array}{cc}-14& 30\\ -18& -32\end{array}\right),\mathrm{Y}=\left(\begin{array}{cc}-18& 20\\ -12& -30\end{array}\right),\mathrm{Z}=\left(\begin{array}{cc}6& 35\\ -21& -15\end{array}\right),\mathrm{T}=\left(\begin{array}{cc}-16& 45\\ -27& -43\end{array}\right),\\ {\mathrm{X}}^2=\left(\begin{array}{cc}-344& -1380\\ 828& 484\end{array}\right),{\mathrm{Y}}^2=\left(\begin{array}{cc}84& -960\\ 576& 660\end{array}\right)\text{,}\\ Z^2=\left(\begin{array}{cc}-699& -315\\ 189& -510\end{array}\right),T^2=\left(\begin{array}{cc}-959& -2655\\ 1593& 634\end{array}\right)\text{,}\end{array}$and the equality (B) is immediately verified.

We can generalize the equations (A), respectively (B) to a certain number n2 of variables / unknowns, as follows:

$x_1^2+x_2^2+\cdots +x_k^2=x_{k+1}^2\left(\mathrm{B}\right)$

From geometrically if (x₁,x₂,…,x_k,x_k+1) $\underset{\mathrm{k}+1}{\underset{⏟}{{\mathrm{N}}^{*}\times {\mathrm{N}}^{*}\times \cdots \times {\mathrm{N}}^{*}}}$ is a solution of the equation (B), then the numbers x₁, x₂, …, x_k represent the dimensions of a right hyperparallelipiped in the space, and the number x_k+1 is the length of its diagonal. (Andreescu & Andrica, 2002)

Following the same reasoning as in the proof of Theorem 5.1.5, the following theorem can be proved:

All integer solutions of the equation (B) are given by the equations:

x₁=2a₁a_k,x₂=2a₂a_k,…x_n-1=2a_k-1a_k,

${\mathrm{x}}_{\mathrm{k}}={\mathrm{a}}^2+{\mathrm{a}}^2+\cdots +{\mathrm{a}}^2{\mathrm{k}}_{-1}{\mathrm{a}}^2, {\mathrm{x}}_{\mathrm{k}+1}={\mathrm{a}}^1+{\mathrm{a}}^2+\cdots +{\mathrm{a}}^{\mathrm{k}-1+{\mathrm{a}}^{\mathrm{k}}}(\mathrm{5.1}.\mathrm{15})$

where a₁, a₂, …, a_k+1 are any integers.

In the ring (M_n(),+,) the equation (B) becomes:

$x_1^2+x_2^2+\cdots +x_k^2=x_{k+1}^2\left(\mathrm{B}\right)$

Equation (B) is solvable in this ring; moreover, we have the following result:

For any matrices A₁, A₂, …, A_kM_n(), such that, for every i, j1,2,…,k,

, (5.1.14)

the following system of k+1 matrices in M_n():

X₁=2A₁A_k,X₂=2A₂A_k,…X_k-1=2A_k-1A_k,

${\mathrm{X}}_{\mathrm{k}}={\mathrm{A}}^2+{\mathrm{A}}^2+\cdots +{\mathrm{A}}^{22}-{\mathrm{A}}^2, {\mathrm{X}}_{\mathrm{k}+1}={\mathrm{A}}^2+{\mathrm{A}}^2+\cdots +{\mathrm{A}}^{22}+{\mathrm{A}}^2(\mathrm{5.1}.\mathrm{15})$

is a solution of the equation (B).

Let be k, k2. Starting from Example 3.7.1 of (Vălcan, 2019), for every i1,2,...,k, we can consider the matrices:

A_i= $\left(\begin{array}{cc}{a}_i& 0\\ c_i& a_i\end{array}\right)$.

Then, for every i, j1,2,…,k:

${\mathrm{A}}_{\mathrm{i}}^2=\left(\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}}^2& 0\\ 2\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{c}}_{\mathrm{i}}& {\mathrm{a}}_{\mathrm{i}}^2\end{array}\right),\text{and}{\mathrm{A}}_{\mathrm{i}}\cdot {\mathrm{A}}_{\mathrm{i}}=\left(\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{j}}& 0\\ {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{c}}_{\mathrm{j}}+{\mathrm{c}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{j}}& {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{j}}\end{array}\right)={\mathrm{A}}_{\mathrm{j}}\cdot {\mathrm{A}}_{\mathrm{i}},$

Now, according to the equalities (5.1.15), for every i= $1,\stackrel{-}{k-1}$, we have:

$\begin{array}{c}{\mathrm{x}}_{\mathrm{i}}=\left(\begin{array}{cc}2\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}& 0\\ 2\cdot \left({\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{c}}_{\mathrm{k}}+{\mathrm{c}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}\right)& 2\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}\end{array}\right),\\ {\mathrm{x}}_{\mathrm{i}}^{\mathrm{2}}=\left(\begin{array}{cc}4\cdot a_i^2\cdot a_k^2& 0\\ 8\cdot a_i\cdot a_k\cdot \left(a_i\cdot c_k+c_i\cdot a_k\right)& 4\cdot a_i^2\cdot a_k^2\end{array}\right),\\ {\mathrm{x}}_{\mathrm{k}}=\left(\begin{array}{cc}{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2& 0\\ 2\cdot \left({\mathrm{a}}_1\cdot {\mathrm{c}}_1+{\mathrm{a}}_2\cdot {\mathrm{c}}_2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{c}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{c}}_{\mathrm{k}}\right)& {\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\end{array}\right)\\ {\mathrm{x}}_{\mathrm{k}+1}=\left(\begin{array}{cc}{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2& 0\\ 2\cdot \left({\mathrm{a}}_1\cdot {\mathrm{c}}_1+{\mathrm{a}}_2\cdot {\mathrm{c}}_2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{c}}_{\mathrm{k}-1}+{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{c}}_{\mathrm{k}}\right)& {\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\end{array}\right)\\ x_k^2=\left(\begin{array}{cc}{\alpha }_k& 0\\ {\beta }_k& {\alpha }_k\end{array}\right)\text{and}{X}_{k+1}^2=\left(\begin{array}{cc}{\gamma }_k& 0\\ {\delta }_k& {\gamma }_k\end{array}\right)\end{array}$

where:

$\begin{array}{c}{\alpha }_{\mathrm{k}}={\left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\right)}^2,\\ {\beta }_{\mathrm{k}}=4\cdot \left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\right)\cdot \left({\mathrm{a}}_1\cdot {\mathrm{c}}_1+{\mathrm{a}}_2\cdot {\mathrm{c}}_2+\cdots {\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{c}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{c}}_{\mathrm{k}}\right),\\ {\gamma }_{\mathrm{k}}={\left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\right)}^2\\ {\beta }_{\mathrm{k}}=4\cdot \left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\right)\cdot \left({\mathrm{a}}_1\cdot {\mathrm{c}}_1+{\mathrm{a}}_2\cdot {\mathrm{c}}_2+\cdots {\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{c}}_{\mathrm{k}-1}+{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{c}}_{\mathrm{k}}\right),\end{array}$

and the equality (B) is immediately verified.

Fie k, k2. Starting from Example 3.8.1 of (Vălcan, 2019), for any i1,2,...,k, we can consider the matrices:

A_i= $\left(\begin{array}{cc}{a}_i& b_i\\ 0& a_i\end{array}\right)$.

Then, for every i, j1,2,…,k:

$A_i^2=\left(\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}}^{\mathrm{2}}& 2\bullet a_i\bullet b_i\\ 0& {\mathrm{a}}_{\mathrm{i}}^{\mathrm{2}}\end{array}\right), \text{and }{A}_i\text{∙}{A}_j=\left(\begin{array}{cc}{\mathrm{a}}_{\mathrm{i}}\bullet a_j& {\mathrm{a}}_{\mathrm{i}}\bullet c_j+c_i\bullet a_{j}0\\ 0& {\mathrm{a}}_{\mathrm{i}}\bullet a_j\end{array}\right)=A_j\bullet A_i,$

Now, according to the equalities (5.1.15), for every i= $i,\stackrel{-}{k-1},$ , we have:

$\begin{array}{c}{\mathrm{x}}_{\mathrm{i}}=\left(\begin{array}{cc}2\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}& 2\cdot \left({\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{b}}_{\mathrm{k}}+{\mathrm{c}}_{\mathrm{i}}\cdot {\mathrm{b}}_{\mathrm{k}}\right)\\ 0& 2\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}\end{array}\right),\\ x_i^2=\left(\begin{array}{cc}4\cdot {\mathrm{a}}_{\mathrm{i}}^2\cdot {\mathrm{a}}_{\mathrm{k}}^2& 8\cdot {\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}\cdot \left({\mathrm{a}}_{\mathrm{i}}\cdot {\mathrm{b}}_{\mathrm{k}}+{\mathrm{b}}_{\mathrm{i}}\cdot {\mathrm{a}}_{\mathrm{k}}\right)\\ 0& 4\cdot {\mathrm{a}}_{\mathrm{i}}^2\cdot {\mathrm{a}}_{\mathrm{k}}^2\end{array}\right)\text{,}\\ {\mathrm{x}}_{\mathrm{d}}=\left(\begin{array}{cc}{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2& 2\cdot \left({\mathrm{a}}_1\cdot {\mathrm{b}}_1+{\mathrm{a}}_2\cdot {\mathrm{b}}_2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{b}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{b}}_k\right)\\ 0& {\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\end{array}\right)\\ {\mathrm{x}}_{\mathrm{k}+1}=\left(\begin{array}{cc}{\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2& 2\cdot \left({\mathrm{a}}_1\cdot {\mathrm{b}}_1+{\mathrm{a}}_2\cdot {\mathrm{b}}_2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{b}}_{\mathrm{k}-1}+{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{b}}_{\mathrm{k}}\right)\\ 0& {\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\end{array}\right),\\ {\mathrm{x}}_{\mathrm{k}}^2=\left(\begin{array}{cc}{\epsilon }_{\mathrm{k}}& {\varphi }_{\mathrm{k}}\\ 0& {\epsilon }_{\mathrm{k}}\end{array}\right)\text{and}{\mathrm{x}}_{\mathrm{k}+1}^2=\left(\begin{array}{cc}{\phi }_{\mathrm{k}}& {\eta }_{\mathrm{k}}\\ 0& {\phi }_{\mathrm{k}}\end{array}\right),\end{array}$

where:

$\begin{array}{c}{\epsilon }_{\mathrm{k}}={\left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\right)}^2\\ {\varphi }_{\mathrm{k}}=4\cdot \left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2-{\mathrm{a}}_{\mathrm{k}}^2\right)\cdot \left({\mathrm{a}}_1\cdot {\mathrm{b}}_1+{\mathrm{a}}_2\cdot {\mathrm{b}}_2+\cdots {\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{b}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{b}}_{\mathrm{k}}\right)\\ {\phi }_{\mathrm{k}}={\left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\right)}^2\\ {\eta }_{\mathrm{k}}=4\cdot \left({\mathrm{a}}_1^2+{\mathrm{a}}_2^2+\cdots +{\mathrm{a}}_{\mathrm{k}-1}^2+{\mathrm{a}}_{\mathrm{k}}^2\right)\cdot \left({\mathrm{a}}_1\cdot {\mathrm{b}}_1+{\mathrm{a}}_2\cdot {\mathrm{b}}_2+\cdots {\mathrm{a}}_{\mathrm{k}-1}\cdot {\mathrm{b}}_{\mathrm{k}-1}+{\mathrm{a}}_{\mathrm{k}}\cdot {\mathrm{b}}_{\mathrm{k}}\right)\end{array}$

and the equality (B) is immediately verified.

Diofantiene quadratic equations

We begin this paragraph by examining the Diophantine equation:

x²+axy+by²=z². (C)

where a and b are given integers. The Pythagorean equation (A) is a special case of this equation which is obtained for:

a=0and b=1.

Regarding equation (C) we have the following result:

All integer solutions of equation (C) are given by the equalities:

x=k(m²b-n²), y=k(2mn-m²a)andz=k(m²b-mna+n²), (5.2.1)

where k, m, n.

A simple calculation shows that the triplets (5.2.1) - defined in the theorem statement, satisfy equation (C). Indeed,

x²=k²(m⁴b²+n⁴-2m²n²b), (5.2.2)

axy=k²(2m³nab-m⁴a²b-2mn³a+m²n²a²), (5.2.3)

by²=k²(4m²n²b-4m³nab+m⁴a²b). (5.2.4)

Now, from the equalities (5.2.2), (5.2.3) and (5.2.4), we obtain:

x²+axy+by²=k²(m⁴b²+m²n²a²+n⁴-2m³nab+2m²n²b-2mn³a)

=z².

Conversely, we will show that all solutions of the equation are of the form above. For this, we observe that equation (C) is equivalent to:

y(ax+by)=(z-x)(z+x). (C)

If

x=z,

then:

y=0orax+by=0. (5.2.5)

In these situations, all triplets of the form (x,0,x) and, respectively $\left(x,-\frac{a\bullet x}{b},x\right)$, with a, b, x and b (ax) are solutions of equation (C). The first sub-case corresponds to the situation when:

ma=2n,

and the second sub-case occurs when:

n=0.

In all other cases the equation (C) is equivalent to:

$\frac{\mathrm{y}}{\mathrm{z}-\mathrm{x}}=\frac{\mathrm{z}+\mathrm{x}}{\mathrm{a}\cdot \mathrm{x}+\mathrm{b}\cdot \mathrm{y}}\stackrel{\text{not.}}{=}\frac{\mathrm{m}}{\mathrm{n}}\left(\mathrm{C}\right)$

where m, n. These equalities lead us to the homogeneous system:

$\left\{\begin{array}{c}\mathrm{m}\cdot \mathrm{x}+\mathrm{n}\cdot \mathrm{y}-\mathrm{m}\cdot \mathrm{z}=0\\ (\mathrm{m}\cdot \mathrm{a}-\mathrm{n})\cdot \mathrm{x}+\mathrm{m}\cdot \mathrm{y}-\mathrm{n}\cdot \mathrm{z}=0\end{array}\right.(\mathrm{5.2}.\mathrm{6})$

whose solutions are:

$\mathrm{x}=\frac{{\mathrm{m}}^2\cdot \mathrm{b}-{\mathrm{n}}^2}{{\mathrm{m}}^2\cdot \mathrm{b}-\mathrm{a}\cdot \mathrm{m}\cdot \mathrm{n}+{\mathrm{n}}^2}\cdot \mathrm{z}\text{ and }\mathrm{y}=\frac{2\cdot \mathrm{m}\cdot \mathrm{n}-{\mathrm{m}}^2\cdot \mathrm{a}}{{\mathrm{m}}^2\cdot \mathrm{b}-\mathrm{a}\cdot \mathrm{m}\cdot \mathrm{n}+{\mathrm{n}}^2}\cdot \mathrm{z}\mathrm{}(\mathrm{5.2}.\mathrm{1})$

Now, we choose:

z=k(m²b-amn+n²), (5.2.1)

with k and obtain the solutions (5.2.1).

In the ring (M_n(),+,) the equation (C) can be of the form:

X²+aXY+bY²=Z², (C)

where a and b are given integers, or may be of the form:

X²+AXY+BY²=Z², (C^(iv))

where A, BM_n() are fixed.

The equalities (5.2.1) entitle us to present the following result:

Equations (C) and (C^(iv)), respectively) are solvable in the ring (M_n(),+,).

Indeed, we notice that if a, b, and A, B, M, NM_n() and satisfy equalities:

AB=BA,AM=MA,MN=NM,AN=NA, BM=MB,BN=NB, (5.2.7)

then the matrices:

X=bM²-N², Y=2MN-aM²andZ=bM²-aMN+N² (5.2.8)

verifies the equality (C), and the matrices:

X₁=BM²-N², Y₁=2MN-AM²andZ₁=BM²-AMN+N² (5.2.9)

verifies the equality (C^(iv)).

Starting also from Example 3.10.1 from (Vălcan, 2019) we can consider the matrices:

A= $\left(\begin{array}{cc}\mathrm{2}& 1\\ 3& \mathrm{2}\end{array}\right)$,B= $\left(\begin{array}{cc}\mathrm{3}& 5\\ 15& \mathrm{3}\end{array}\right)$, M= $\left(\begin{array}{cc}\mathrm{4}& 15\\ 45& \mathrm{4}\end{array}\right)$,andN= $\left(\begin{array}{cc}\mathrm{5}& 2\\ 6& \mathrm{5}\end{array}\right)$.

Then:

M²= $\left(\begin{array}{cc}\mathrm{691}& 120\\ 360& \mathrm{691}\end{array}\right)$, andN²= $\left(\begin{array}{cc}\mathrm{37}& 20\\ 60& \mathrm{37}\end{array}\right)$,

AB= $\left(\begin{array}{cc}\mathrm{21}& 13\\ 39& \mathrm{21}\end{array}\right)$=BA,AM= $\left(\begin{array}{cc}\mathrm{53}& 34\\ 102& \mathrm{53}\end{array}\right)$=MA, MN= $\left(\begin{array}{cc}\mathrm{110}& 83\\ 249& \mathrm{110}\end{array}\right)$=NM,

AN= $\left(\begin{array}{cc}\mathrm{16}& 9\\ 27& \mathrm{16}\end{array}\right)$=NA,BM= $\left(\begin{array}{cc}\mathrm{237}& 65\\ 195& \mathrm{237}\end{array}\right)$=MA, BN= $\left(\begin{array}{cc}\mathrm{45}& 31\\ 93& \mathrm{45}\end{array}\right)$=BN.

Now, according to equality (5.2.8), for:

b=3anda=2,

we have:

X= $\left(\begin{array}{cc}\mathrm{2036}& 340\\ 1020& \mathrm{2036}\end{array}\right)$,Y= $\left(\begin{array}{cc}-\mathrm{1162}& -74\\ -222& -\mathrm{1162}\end{array}\right)$andZ= $\left(\begin{array}{cc}\mathrm{1890}& 214\\ 642& \mathrm{1890}\end{array}\right)$,

X²= $\left(\begin{array}{cc}\mathrm{4492096}& 1384480\\ 4153440& \mathrm{4492096}\end{array}\right)$,XY= $\left(\begin{array}{cc}-\mathrm{2441312}& -545744\\ -1637232& -\mathrm{2441312}\end{array}\right)$=YX,

Y²= $\left(\begin{array}{cc}\mathrm{1366672}& 171976\\ 515928& \mathrm{1366672}\end{array}\right)$, Z²= $\left(\begin{array}{cc}\mathrm{3709488}& 808920\\ 2426760& \mathrm{3709488}\end{array}\right),$

and the equality (C), for the case considered, is verified immediately:

X²+2XY+3Y²= $\left(\begin{array}{cc}\mathrm{4492096}& 1384480\\ 4153440& \mathrm{4492096}\end{array}\right)$+ $\left(\begin{array}{cc}-\mathrm{4882624}& -1091488\\ -3274464& -\mathrm{4882624}\end{array}\right)$+ $\left(\begin{array}{cc}\mathrm{4100016}& 515928\\ 1547784& \mathrm{4100016}\end{array}\right)$

= $\left(\begin{array}{cc}\mathrm{3709488}& 808920\\ 2426760& \mathrm{3709488}\end{array}\right)$=Z².

On the other hand, according to the equalities (5.2.9), we obtain:

X₁= $\left(\begin{array}{cc}\mathrm{3836}& 3795\\ 11385& \mathrm{3836}\end{array}\right)$,Y₁= $\left(\begin{array}{cc}-\mathrm{1522}& -765\\ -2295& -\mathrm{1522}\end{array}\right)$andZ₁= $\left(\begin{array}{cc}\mathrm{3441}& 3559\\ 10677& \mathrm{3441}\end{array}\right)$,

$x_1^2=\left(\begin{array}{cc}\mathrm{57920971}& 29115240\\ 87345720& \mathrm{57920971}\end{array}\right)$,X₁Y₁= $\left(\begin{array}{cc}-\mathrm{14547917}& -8710530\\ -26131590& -\mathrm{14547917}\end{array}\right)$=Y₁X₁,

$Y_1^2=\left(\begin{array}{cc}\mathrm{4072159}& 2328660\\ 6985980& \mathrm{4072159}\end{array}\right)$, $Z_1^2\left(\begin{array}{cc}\mathrm{49839924}& 24493038\\ 73479114& \mathrm{49839924}\end{array}\right)$,

and the equality (C^(iv)), for the case considered, is verified immediately:

$X_1^2+A\bullet X_1\bullet Y_1+B\bullet Y_1^2=\left(\begin{array}{cc}\mathrm{57920971}& 29115240\\ 87345720& \mathrm{57920971}\end{array}\right)-\left(\begin{array}{cc}\mathrm{55227424}& 31968977\\ 95906931& \mathrm{55227424}\end{array}\right)+\left(\begin{array}{cc}\mathrm{47146377}& 27346775\\ 82040325& \mathrm{47146377}\end{array}\right)=\left(\begin{array}{cc}\mathrm{49839924}& 24493038\\ 73479114& \mathrm{49839924}\end{array}\right)=Z_1^2$

A particular case of equation (C) we obtain for:

b=a²,

when equation (C) becomes:

x²+axy+a²y²=z². (D)

From Theorem 5.2.1 we obtain immediately:

The integer solutions of equation (D) are:

x=k(m²a²-n²), y=k(2mn-m²a)andz=k(m²a²-mna+n²), (5.2.10)

where k, m, n.

In the ring (M_n(),+,) the equation (D) can be of the form:

X²+aXY+a²Y²=Z², (D)

where a is an integer, given, or can be of the form:

X²+AXY+A²Y²=Z², (D)

where AM_n() is fixed.

The equalities (5.2.10) entitle us to present the following result:

Equations (D) and (D), respectively, are solvable in the ring (M_n(),+,).

Indeed, we observe that if, and A, M, NM_n() and satisfy the equalities:

AM=MA,MN=NM,AN=NA, (5.2.11)

then the matrices:

X=a²M²-N², Y=2MN-aM²andZ=a²M²-aMN+N² (5.2.12)

verifies the equality (D) and the matrices:

X₁=A²M²-N², Y₁=2MN-AM²andZ₁=A²M²-AMN+N² (5.2.13)

verifies the equality (D).

We also consider the matrices here:

A= $\left(\begin{array}{cc}2& 1\\ 3& \mathrm{2}\end{array}\right)$,M= $\left(\begin{array}{cc}4& 15\\ 45& \mathrm{4}\end{array}\right)$, andN= $\left(\begin{array}{cc}5& 2\\ 6& \mathrm{5}\end{array}\right)$.

Then:

A²= $\left(\begin{array}{cc}7& 4\\ 12& \mathrm{7}\end{array}\right)$,M²= $\left(\begin{array}{cc}691& 120\\ 360& \mathrm{691}\end{array}\right)$ andN²= $\left(\begin{array}{cc}37& 20\\ 60& \mathrm{37}\end{array}\right)$,

AM= $\left(\begin{array}{cc}53& 34\\ 102& \mathrm{53}\end{array}\right)$=MA, MN= $\left(\begin{array}{cc}110& 83\\ 249& \mathrm{110}\end{array}\right)$=NM,AN= $\left(\begin{array}{cc}16& 9\\ 27& \mathrm{16}\end{array}\right)$=NA.

Now, according to the equalities (5.2.12), for:

a=2,

we have:

X= $\left(\begin{array}{cc}2727& 460\\ 1380& \mathrm{2727}\end{array}\right)$,Y= $\left(\begin{array}{cc}-1162& -74\\ -222& -\mathrm{1162}\end{array}\right)$,andZ= $\left(\begin{array}{cc}2581& 334\\ 1002& \mathrm{2581}\end{array}\right)$,

X²= $\left(\begin{array}{cc}8071329& 2508840\\ 526520& \mathrm{8071329}\end{array}\right)$,XY= $\left(\begin{array}{cc}-3270894& -736318\\ -2208954& -\mathrm{3270894}\end{array}\right)$=YX,

Y²= $\left(\begin{array}{cc}1366672& 171976\\ 515928& \mathrm{1366672}\end{array}\right)$,Z²= $\left(\begin{array}{cc}6996229& 1724108\\ 5172324& \mathrm{6996229}\end{array}\right)$,

and equality (D), for the case in question, is verified immediately:

X²+2XY+4Y²= $\left(\begin{array}{cc}8071329& 2508840\\ 7526520& \mathrm{8071329}\end{array}\right)-\left(\begin{array}{cc}6541788& 1472636\\ 4417908& \mathrm{6541788}\end{array}\right)+\left(\begin{array}{cc}5466688& 687904\\ 2063712& \mathrm{5466688}\end{array}\right)=\left(\begin{array}{cc}6996229& 1724108\\ 5172324& \mathrm{696229}\end{array}\right)=Z^2$

On the other hand, according to the equalities (5.2.13), we obtain:

X₁= $\left(\begin{array}{cc}6240& 3584\\ 10752& \mathrm{6240}\end{array}\right)$,Y₁= $\left(\begin{array}{cc}-1522& -765\\ -2295& -\mathrm{1522}\end{array}\right)$,andZ₁= $\left(\begin{array}{cc}5845& 3348\\ 10044& \mathrm{5845}\end{array}\right)$, $X_1^2=\left(\begin{array}{cc}77472768& 44728320\\ 134184960& \mathrm{77472768}\end{array}\right)$,X₁Y₁= $\left(\begin{array}{cc}-17722560& -10228448\\ -30685344& -\mathrm{17722560}\end{array}\right)$=Y₁X₁, $Y_1^2=\left(\begin{array}{cc}4072159& 2328660\\ 6985980& \mathrm{4072159}\end{array}\right)$, $Z_1^2=\left(\begin{array}{cc}67791337& 39138120\\ 117414360& \mathrm{67791337}\end{array}\right)$,

and the equality (D), for the case considered, is verified immediately:

$X_1^2+A\bullet X_1\bullet Y_1+A^2\bullet Y_1^2=\left(\begin{array}{cc}77472768& 44728320\\ 134184960& \mathrm{77472768}\end{array}\right)-\left(\begin{array}{cc}\mathrm{66130464}& 38179456\\ 114538368& \mathrm{66130464}\end{array}\right)+\left(\begin{array}{cc}\mathrm{56449033}& 32589256\\ 97767768& \mathrm{56449033}\end{array}\right)=\left(\begin{array}{cc}\mathrm{67791337}& 39138120\\ 117414360& \mathrm{67791337}\end{array}\right)=Z_1^2$

The following two particular cases of equation (D) are interesting:

a=1. In this case equation (D) becomes:

x²+xy+y²=z². (D)

From the equalities (5.2.10) it results that its non-zero natural solutions are given by:

x=k(m²-n²), y=k(2mn-m²) and z=k(m²-mn+n²), (5.2.14)

where k, m, n, and mn $\frac{m}{2}$

The solutions (5.2.14) give all triplets (x, y, z) of nonzero natural numbers which are the lengths of the sides of a triangle with the angle opposite to the side of length z equal to 120. (Andreescu & Andrica, 2002)

a=-1. In this case equation (D) becomes:

x²-xy+y²=z². (D^(iv))

From the same equalities (5.2.10) it results that its non-zero natural solutions are given by the equalities:

x=k(m²-n²), y=k(2mn+m²) and z=k(m²+mn+n²), (5.2.15)

where k, m, n, and mn.

The solutions (5.2.15) characterize all triplets of non-zero natural numbers (x, y, z) which are the lengths of the sides of a triangle having the angle opposite to the side of length z equal to 60 (Andreescu & Andrica, 2002)

In the ring (M_n(),+,) the equations (D) and (D^(iv)) respectively become:

X²+XY+Y²=Z² (D^(v))

respectively:

X²-XY+Y²=Z². (D^(vi))

The results obtained above entitle us to present:

Equations (D^(v)) and (D^(vi)), respectively, are solvable in the ring (M_n(),+,).

Indeed, we observe that if M, NM_n() and satisfy the equality:

MN=NM, (5.2.16)

then the matrices:

X=M²-N², Y=2MN-M²andZ=M²-MN+N² (5.2.17)

verifies the equality (D^(v)), and the matrices:

X₁=M²-N², Y₁=2MN+M²andZ₁=M²+MN+N² (5.2.18)

verifies the equality (D^(vi)).

We can consider the matrices in Examples 5.2.3:

M= $\left(\begin{array}{cc}\mathrm{4}& 15\\ 45& \mathrm{4}\end{array}\right)$, andN= $\left(\begin{array}{cc}\mathrm{5}& 2\\ 6& \mathrm{5}\end{array}\right)$.

Then:

M²= $\left(\begin{array}{cc}\mathrm{691}& 120\\ 360& \mathrm{691}\end{array}\right)$, N²= $\left(\begin{array}{cc}\mathrm{37}& 20\\ 60& \mathrm{37}\end{array}\right)$andMN= $\left(\begin{array}{cc}\mathrm{110}& 83\\ 249& \mathrm{110}\end{array}\right)$=NM.

Now, according to the equalities (5.2.17), we have:

X= $\left(\begin{array}{cc}\mathrm{654}& 100\\ 300& \mathrm{654}\end{array}\right)$,Y= $\left(\begin{array}{cc}-\mathrm{471}& 46\\ 138& -\mathrm{471}\end{array}\right)$,andZ= $\left(\begin{array}{cc}\mathrm{618}& 57\\ 171& \mathrm{618}\end{array}\right)$,

X²= $\left(\begin{array}{cc}\mathrm{457716}& 130800\\ 392400& \mathrm{457716}\end{array}\right)$,XY= $\left(\begin{array}{cc}-\mathrm{294234}& -17016\\ -51048& -\mathrm{294234}\end{array}\right)$=YX,

Y²= $\left(\begin{array}{cc}\mathrm{228189}& -43332\\ -129996& \mathrm{228189}\end{array}\right)$, Z²= $\left(\begin{array}{cc}\mathrm{391671}& 70452\\ 211356& \mathrm{391671}\end{array}\right)$,

and the equality (D^(v)), is verified immediately:

X²+XY+Y²= $\left(\begin{array}{cc}\mathrm{457716}& 130800\\ 392400& \mathrm{457716}\end{array}\right)$+ $\left(\begin{array}{cc}-\mathrm{294234}& -17016\\ -51048& -\mathrm{294234}\end{array}\right)$+ $\left(\begin{array}{cc}\mathrm{228189}& -43332\\ -129996& \mathrm{228189}\end{array}\right)$

= $\left(\begin{array}{cc}\mathrm{391671}& 70452\\ 211356& \mathrm{391671}\end{array}\right)$

=Z².

On the other hand, according to the equalities (5.2.9), we obtain:

X₁= $\left(\begin{array}{cc}\mathrm{654}& 100\\ 300& \mathrm{654}\end{array}\right)$,Y₁= $\left(\begin{array}{cc}\mathrm{911}& 286\\ 858& \mathrm{911}\end{array}\right)$,andZ₁= $\left(\begin{array}{cc}\mathrm{838}& 223\\ 669& \mathrm{838}\end{array}\right)$,

$X_1^2=\left(\begin{array}{cc}\mathrm{457716}& 130800\\ 392400& \mathrm{457716}\end{array}\right)$,X₁Y₁= $\left(\begin{array}{cc}\mathrm{681594}& 278144\\ 834432& \mathrm{681594}\end{array}\right)$=Y₁X₁,

$Y_1^2=\left(\begin{array}{cc}\mathrm{1075309}& 521092\\ 1563276& \mathrm{1075309}\end{array}\right)$, $Z_1^2=\left(\begin{array}{cc}\mathrm{851431}& 373748\\ 1121244& \mathrm{851431}\end{array}\right)$,

and the equality (C^(iv)), for the case considered, is verified immediately:

$\begin{array}{c}{\mathrm{X}}_1^2-{\mathrm{X}}_1\cdot {\mathrm{Y}}_1+{\mathrm{Y}}_1^2=\left(\begin{array}{cc}457716& 130800\\ 392400& 457716\end{array}\right)\\ =\left(\begin{array}{cc}851431& 373748\\ 1121244& 851431\end{array}\right)\\ ={\left(\begin{array}{cc}681594& 278144\\ 834432& 681594\end{array}\right)}_{+}\left(\begin{array}{cc}1075309& 521092\\ 1563276& 1075309\end{array}\right)\\ ={\mathrm{Z}}_1^2\end{array}$

The following remarks are required here:

Theorem 5.2.1 shows how to solve the following third degree Diophantine equation:

. (E)

Its general solution is (x,y,z,t), where:

t=a,

with a and x, y, z are given by the equalities (5.2.1) - for b=1.

Combining the results proved above we can solve the Diophantine equation:

. (F)

Its solutions are (x,y,z,t,u,v), where:

u=aand v=b,

with a, b, and x, y, z are given by the equalities (5.2.1).

In the ring (M_n(),+,) the equations (E) and (F) respectively become:

X²+XYT+Y²=Z² (E)

respectively:

X²+UXY+VY²=Z². (F)

And at the end of this paragraph we can present:

Equations (E) and (F), respectively, are solvable in the ring (M_n(),+,).

Indeed, we observe that if A, B, M, NM_n() and satisfy the equalities:

AM=MA,MN=NM,AN=NA, (5.2.11)

BM=MB,MN=NM,BN=NB, (5.2.11)

and moreover:

AB=BA, (5.2.16) then, according to the proof of Theorem 5.2.2, the matrices:

X=M²-N², Y=2MN-AM², Z=M²-AMN+N² and T=A (5.2.13)

verifies the equality (E), and the matrices:

X₁=BM²-N², Y₁=2MN-AM², Z₁=BM²-AMN+N²,

U=A andV=B (5.2.9)

verifies the equality (F).