Quaternionic Algebra Reformulation and Proof of Fermat’s Last Theorem for All Exponents Greater than 2

2.1. Quaternionic Algebra and Complexification

Quaternions form a four-dimensional normed division algebra over the real numbers, denoted ℍ, consisting of one real unit element 1 and three imaginary units e₁, e₂, and e₃. These imaginary units satisfy the following anti-commutative multiplication rules

$${e_1}^2={e_2}^2={e_3}^2=-1,$$

$$e_1{e}_2=e_3,e_2{e}_3=e_1,e_3{e}_1=e_2,$$

$$e_i{e}_j+e_{j}e{e}_i=0,for i\ne j.$$

(1)

2.2. Pythagorean Theorem n = 2

We define $\mathit{A}=a{e}_1+b e_2+\mathrm{i} c{e}_3$ ∈ Q_C and a, b, c ∈ Z. One can show

$${\mathrm{A}}^2=-({\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2)=-{\Vert \mathrm{A}\Vert }^2\in \mathrm{Z}$$

(2A)

and

$$\mathit{exp}\left(i2\pi A\right)={{\displaystyle \sum }}_{k=0}^{\infty }{\left(i2\pi A\right)}^k/k!=\mathit{cos}\left(2\pi \Vert A\Vert \right)+iA\mathit{sinc}\left(2\pi \Vert A\Vert \right)\in \mathrm{C}$$

(2B)

where the sinc-function$\mathit{sinc}\left(x\right)\equiv \mathit{sin}\left(x\right)/x$ approaches 1 if x approaches 1.

For $\mathit{exp}\left(i2\pi A\right)=1$, from $\mathit{cos}\left(2\pi \Vert A\Vert \right)=1$ and $\mathit{sinc}\left(2\pi \Vert A\Vert \right)$ =0, one obtains $\Vert A\Vert =\sqrt{a_2+b^2-c^2}=m$ ∈ Z, and if m $\ne$0, one obtains $a^2+b^2=c^2+m^2$ , which can be satisfied by infinite integer sets of a, b, c, and m. If m $=0,$ one obtains $a_2+b^2=c^2$ which is the standard Pythagorean theorem, known to possess numerous integer solutions.

2.3. FLT Proof for n = 4

Assume $A=a^2{e}_1+b^2{e}_{2r}+{\omega }^2{c}^2{e}_3,\omega =\mathit{exp}\left(i\pi /4\right)$ ∈ Q_c, one obtains

$$A^2=-\left(a^4+b^4-c^4\right)=-{\Vert A\Vert }^2\in \mathrm{Z}$$

(3A)

and

$$\begin{gathered} \mathit{exp}\left(i2\pi A\right)={\displaystyle \sum }_{k=0}^{\infty }{\left(i2\pi A\right)}^k/k!=\mathit{cos}\left(2\pi \Vert A\Vert \right)+iA\mathit{sinc}\left(2\pi \Vert A\Vert \right),\Vert A\Vert \\ =\sqrt{a^4+b^4-c^4} \end{gathered}$$

(3B)

For $\mathit{exp}\left(i2\pi A\right)=1$, from $\mathit{cos}\left(2\pi \Vert A\Vert \right)=1$ and $A\mathit{sinc}\left(2\pi \Vert A\Vert \right)$ =0, one obtains $\Vert A\Vert =\sqrt{a_2+b^2-c^2}=m$ ∈ Z, and if m $\ne$0, one obtains $a^4+b^4=c^4+m^2$. For FLT with a power of 4, one has m = 0, one has$\Vert A\Vert =0$, which leads to $\mathit{sinc}\left(2\pi \Vert A\Vert \right)\to 1$ and $A=0$, i.e., $a^2{e}_1+b^2{e}_{2r}+{\omega }^2{c}^2{e}_3=0$.

This vanishing quaternion can be satisfied only if a = b = c = 0. Thus, this proves FLT for n = 4. For m different from 0, one has $a^4+b^4-c^4=m^2$ which is unrelated to FLT for n = 4, not need to be explored here.

2.4. FLT Proof n=2k

Assume $A=a^k{e}_1+b^k{e}_{2r}+{\omega }^k{c}^k{e}_3,\omega =\mathit{exp}\left(i\pi /2k\right)$ ∈ Q_c, one obtains

$$\begin{gathered} A^2=-\left(a^{2k}+b^{2k}-c^{2k}\right)=-{\Vert A\Vert }^2 \\ \Vert A\Vert =\sqrt{a^{2k}+b^{2k}-c^{2k}},\in \mathrm{R} \end{gathered}$$

(4A)

and

$$\mathit{exp}\left(i2\pi A\right)={{\displaystyle \sum }}_{k=0}^{\infty }{\left(i2\pi A\right)}^k/k!=\mathit{cos}\left(2\pi \Vert A\Vert \right)+iA\mathit{sinc}\left(2\pi \Vert A\Vert \right)$$

(4B)

For $\mathit{exp}\left(i2\pi A\right)=1$, from $\mathit{cos}\left(2\pi \Vert A\Vert \right)=1$ and $A\mathit{sinc}\left(2\pi \Vert A\Vert \right)$ =0. Like the n=4 case, to meet these constraints, one must have $\Vert A\Vert =\sqrt{a^{2k}+b^{2k}-c^{2k}}=m$ is an integer. For FLT with a power of 2k, one has m = 0, one has$\Vert A\Vert =0$, which leads to $\mathit{sinc}\left(2\pi \Vert A\Vert \right)\to 1$ and $A=0$. Because$A=a^k{e}_1+b^k{e}_{2r}+i^k{c}^k{e}_3=0$, this leads to a = b = c = 0, thus, proving FLT for 2n.

2.5. FLT Proof n = 3

We define $P=a{e}_1+b{e}_2+\omega c{e}_3$and$Q=a^2{e}_1+b^2{e}_2+{\omega }^2{c}^2{e}_3,$where $\omega =\mathit{exp}\left(i\pi /3\right)$, and R = P ⋅ Q. This clan be shown that results in a scalar part and three imaginary components along e₁, e₂, e₃, namely,

$$R=D+A{e}_1+B{e}_{2r}+C{e}_3$$

(5A)

where

$$\begin{gathered} D=-\left(a^3+b^3-c^3\right) \\ A=b\omega \left(b-\omega c\right) \\ B=a\omega \left(\omega c-a\right) \\ C=ab\left(a-b\right) \end{gathered}$$

(5B)

With FLT for n = 3, one has $-D=a^3+b^3-c^3=0$ and $\Vert R\Vert ={\sqrt{-R}}^2=\sqrt{A^2+B^2+C^2}$. For $\mathit{exp}\left(-i2\pi R\right)=1$, from $\mathit{cos}\left(2\pi \Vert R\Vert \right)=1$ and $R\mathit{sinc}\left(2\pi \Vert R\Vert \right)$ =0. Like previous analyses, to meet these constraints, one must have$\sqrt{A^2+B^2+C^2}=m$ where m is an integer. If m =0, one must have R = 0. Because $R=A{e}_1+B{e}_2+C{e}_3$, it leads to A = B = C = 0. To satisfy such constraints, one must have a=b=c=0, thus proving FLT. If m $\ne$0, one obtains $A^2+B^2+C^2=m^2$, i.e., $b^2{\omega }^2{\left(b-\omega c\right)}^2+a^2{\omega }^2{\left(\omega c-a\right)}^2+a^2{b}^2{\left(a-b\right)}^2=m^2$. Because $\omega =\mathit{exp}\left(i\pi /3\right)$ , together with $-D=a^3+b^3-c^3=0$, these constraints lead to a = b = c = m = 0, thus, proving FLT for n = 3.

2.6. FLT Proof 2k+1

We define $P=a^{k+1}{e}_1+b^{k+1}{e}_2+{\omega }^{k+1}{c}^{k+1}{e}_3$and$Q=a^k{e}_1+b^k{e}_2+{\omega }^k{c}^k{e}_3,$where $\omega =\mathit{exp}\left(i\pi /\left(2k+1\right)\right)$, and R = P ⋅ Q. Tit clan be shown that result in a scalar part and three imaginary components along e₁, e₂, e₃, namely,

$$R=D+A{e}_1+B{e}_{2r}+C{e}_3$$

(6A)

where

$$\begin{gathered} D=-\left(a^{2k+1}+b^{2k+1}-c^{2k+1}\right) \\ A=b^k{\omega }^k\left(b-\omega c\right) \\ B=a^k{\omega }^k\left(\omega c-a\right) \\ C=a^k{b}^k\left(a-b\right). \end{gathered}$$

(6B)

With FLT for n = 2k+1, one has $-D=a^{2k+1}+b^{2k+1}-c^{2k+1}=0$ and $\Vert R\Vert ={\sqrt{-R}}^2=\sqrt{A^2+B^2+C^2}$.

For $\mathit{exp}\left(-i2\pi R\right)=1$, from $\mathit{cos}\left(2\pi \Vert R\Vert \right)=1$ and $R\mathit{sinc}\left(2\pi \Vert R\Vert \right)$ =0. Like the previous analysis, to meet these constraints, one must have$\sqrt{A^2+B^2+C^2}=m$ where m is an integer. If m =0, one must have R = 0. Because $R=A{e}_1+B{e}_2+C{e}_3$, it leads to A = B = C = 0. To satisfy such constraints, one must have a=b=c=0, thus proves FLT. If m $\ne$0, one obtains $A^2+B^2+C^2=m^2$, i.e., $b^2{\omega }^{2k}{\left(b-\omega c\right)}^2+a^{2k}{\omega }^2{\left(\omega c-a\right)}^{k2}+a^{2k}{b}^{2k}{\left(a-b\right)}^2=m^2$. Because $\omega =\mathit{exp}\left(i\pi /3\right)$ , together with $-D=a^{2k+1}+b^{2k+1}-c^{2k+1}=0$these constraints leadl to a = b = c = m= 0, thus, proving FLT for n = 2k+1.