The article shows the application of some simple geometric methods to Fermat's last theorem. In particular, the application of the method of n-lines to the solution of this problem is shown [1,2,5], that is, the division of the sides of a triangle is proportional to the n-th powers of the adjacent sides. As is known before Andrew Wiles [6], many people tried to prove this theorem, as a result of which many interesting results were obtained [7], which contributed to the development of mathematics in general. Here, as in many authors, an attempt is made to prove by contradiction. That is, the attempt begins with the assumption that there are positive integers a, b, c satisfying the equality $a^n+b^n=c^n$, where n (n ˃ 2) is an integer, in order to show that this is impossible and leads to a contradiction.

It is stated that the proof of the last Fermat's theorem, which has not been solved for nearly three centuries, is self-evident only when the power is 4 (Edwards, 1980). Some also note that although Fermat did not write the proof of the theorem, he knew the proof himself (Poulkas, 2020). About a century later, Euler was able to prove the theorem for the case where the power was 3, and two centuries later, Dirichlet and Legendre proved the theorem for the case when the power was 5. Lame demonstrated the theorem for the case where the power was 7. Wels announced that he had proved the theorem in 1993, but after the presentation, it was revealed that there was a flaw in the proof. Wiles presented the final proof of the theorem two years later, in 1995, together with his former student Taylor (Taylor, Wiles, 1995). Wiles and Taylor's proof is very long, some 130 pages. Considering this, later, many mathematicians worked on the existence of more straightforward alternative ways of proof. As an example, we will mention several approaches that have been proposed in recent times. Agofontsev (2012) and Gevorkyan (2020a, 2020b) investigated the theorem for certain cases and obtained some general conclusions. Poulkas (2020) states his results regarding the proof of both LFT and his proposed Generalized Fermat's Theorem (the case when there are n terms in the sum) using number theory.

Although certain tools of mathematical analysis and number theory are used to prove the theorem in this article, the geometrical approach is the main focus. A special family of curves called Elba curves is introduced to facilitate the proof.

In this article, many interesting results have been obtained that have not been previously published.

Fermat's Last Theorem states that for any natural number n>2 the equation $a^n+b^n=c^n$ has no solutions in non-zero integers a, b, c.

In the article [3] “Towards the proof of Fermat's theorem”, we showed that the solution of this problem is adequate to finding ΔABC with integer sides a, b, c, for which

$a^n+b^n=c^n$, where natural n more than two (n > 2).

Consider the problem of dividing the sides of a triangle proportionally n-th powers of adjacent sides [1,2,5].

Let ΔABC be given with corresponding sides a, b, c. Let's draw a straight-line CD from vertex C to side AB.

If:

All three n-lines intersect at one point, which is easily proved by the Ceva’s theorem.

Consider the following problem. Let the lengths of the sides of the triangle be integers. Let's determine for what values of n the points of intersection of the lines of n - straight lines lie on the midline.

As is known, the point “k” belongs to the n-line CD if and only if the distances from “k” to sides a and b are proportional to the (n-1)-th powers of these sides (see Figure 1), i.e.

For clarity, we present the proof of this statement.

Let the angles at the vertex C be ${\gamma }_1$and ${\gamma }_2$Then

Let's carry out the following transformations:

Then

On the other hand, S_ΔACD=1/2·h·AD, S_ΔBCD=1/2·h·BD, where h is the height drawn from vertex C to side AB. Then, using the relations $\frac{AD}{DB}=\frac{b^n}{a^n}$, we get

And now let the point K be the point of intersection of n-lines drawn from different vertices, and the points M, L, N based on the perpendiculars drawn to the corresponding sides a, b, c. Using the previous formula, we can get the following formula

After doing some transformations, we get:

Let this point lie on the midline parallel to side c. Then

What does the last formula mean? For n=0 it has 1+1≠1, i.e. the point of intersection of the medians cannot lie on the midline. For n=1 we have a+b=c, which contradicts the triangle inequality. That is, the point of intersection of the bisectors also cannot lie on the midline.

For n=2 we get a right triangle because

As is known, Only the Simedians of the rectangular triangle intersect in the middle of the height and $\frac{b^2}{a^2}=\frac{AD}{DB}$, where $CD=h=\frac{ab}{c}$If we assume that the point K is in the middle of the CD-line of n-lines, then we can write the following equality: (see Figure 2).

Figure 2 xxx.

Where

On the other hand, by the property of the n-straight lines

Given the formulas (14) in (13), we get: $a·\frac{a^{n-1}}{c^{n-1}}+b·\frac{b^{n-1}}{c^{n-1}}=c$ or $a^n+b^n=c^n$It should be noted that the same result can be obtained by comparing the formulas:

and$a^n+b^n=c^n$.

As you can see with

we get: $a^n+b^n=c^n$And for all values n formulas (16) are true? With n = 2, as we have already noted, a rectangular triangle is obtained.

And formulas $cos\alpha =\frac{b^{2-1}}{c^{2-1}}=\frac{b}{c},cos\beta =cos\alpha =\frac{a^{2-1}}{c^{2-1}}=\frac{a}{c}$ are true.

To get a unambiguous answer to this question with n > 2, we first get formulas for N-straight lengths. To do this, we use the Stuart formula.

Let $AD=m_1=k{b}^n,DB=m_{1}k{a}^n$Then according to the Stuart formula (see Figure 3)

On the other hand, after the substitution of values for AD and DB we have:

By substituting (19) in (18) and conducting some transformations, we have:

Similar formulas can be obtained for the other two Simedians:

Let us show that these formulas can be used to obtain formulas for the median, bisector and symmedian of the triangle ∆ABC

For n=0, from formula (20) we obtain the median formula:

For n=1, we get the bisector formula:

For n=2, we get the formula for the symmedian:

The formula for the length of n-lines in a triangle in which $a^n+b^n=c^n$can be written as follows:

And now we will show that in a triangle in which the n-line CD=L_c is drawn, where the equality $a^n+b^n=c^n$ holds, formulas (16) are true only for n=2.

Indeed, this can be obtained if we consider ∆ACD and ∆BCD

First consider the triangle ∆ACD

For the triangle ∆ACD, we write the cosine theorem and instead of cosα, we substitute the value $cos\alpha =\frac{b^{n-1}}{c^{n-1}}$ from formula (19) substituting $a^n+b^n=c^n$we get $AD=\frac{c{b}^n}{c^n}=\frac{b^n}{c^{n-1}}, DB=\frac{c{a}^n}{c^n}=\frac{a^n}{c^{n-1}}$

The cosine theorem has the following form:

Substituting the values of AD and cosα, we obtain:

Similarly, from the triangle ∆ACB we have:

Equating equations (29) and (30), we obtain:

The left side of the last equation is divisible by c, then the right side must also be divisible by c, which contradicts this condition.

Hence n = 2.

And now we will show geometrically that the equalities аⁿ + bⁿ = cⁿ and

acosβ+bcosα = c are satisfied for cosβ = $\frac{a^{n-1}}{c^{n-1}},$ cosα = $\frac{b^{n-1}}{c^{n-1}}$ only for n = 2, and at the same time the line of n-straight lines СН ⊥ АВ.

We have already shown that to satisfy the equality $a^{n }+b^{n }=c^n,$ where a, b, c and n are natural numbers, the following equalities must be satisfied:

That is: a$\frac{x}{z}+b\frac{y}{z}$ = c⇒ $a\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\mathsf{\beta }}_1}{\mathrm{s}\mathrm{i}\mathrm{n}{\mathsf{\beta }}_2}+b\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_1}{\mathrm{s}\mathrm{i}\mathrm{n}{\mathsf{\alpha }}_2}=\mathrm{с}\Rightarrow$ a$\frac{a^{n-1}}{c^{n-1}}+b\frac{b^{n-1}}{c^{n-1}}=c\Rightarrow a^n+b^n=c^n$ On the other hand, all these formulas are written for the triangle ABC with sides a, b, c and corresponding angles α, β, γ, for which the equality: acosβ + bcosα=c

Comparing the formulas, the last formulas can be written: cosβ = $\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\mathsf{\beta }}_1}{\mathrm{s}\mathrm{i}\mathrm{n}{\mathsf{\beta }}_2},$ cosα = $\frac{\mathrm{c}\mathrm{o}\mathrm{s}{\mathsf{\alpha }}_1}{\mathrm{c}\mathrm{o}\mathrm{s}{\mathsf{\alpha }}_2}$Let's transform these formulas. All trigonometric expressions obtained for angles α, α₁ α₂ will be true for angles β, β₁, β₂. Thus we get:

To visualize the application of the last equality, let's draw a triangle ABC' (ΔABC'=ΔABC), which is symmetrical to the triangle ABC with respect to the side AB (see Figure 5)

As we can see from the previous formula, the ratio of the sines should be $\frac{1}{3}$.

Then $\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_1}{\mathrm{s}\mathrm{i}\mathrm{n}(2{\alpha }_2+{\alpha }_1)}=\frac{h_1}{h_2}$ =$\frac{1}{3}$ Since $\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_1}{\mathrm{s}\mathrm{i}\mathrm{n}(2{\alpha }_2+{\alpha }_1)}=\frac{h_1}{h_2}$ = $\frac{\frac{1}{2}b{h}_1}{\frac{1}{2}b{h}_2}=\frac{B}{M,}$ and only when ВВ' ⊥ АС

СМ=МН=НМ'=М'С', then

$\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_1}{\mathrm{s}\mathrm{i}\mathrm{n}(2{\alpha }_2+{\alpha }_1)}=\frac{h_1}{h_2}$ =$\frac{ВМ}{В\text{'} М\text{'}}=\frac{1}{3} ,$ only then the line of n straight lines is perpendicular to the base (for n = 2) (see Figure 6).

Let us show that for n >2 the ratio of sines is not equal to$\frac{1}{3}$.

Indeed, $\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\alpha }_1}{\mathrm{s}\mathrm{i}\mathrm{n}(2{\alpha }_2+{\alpha }_1)}=\frac{\frac{h_1}{AM}}{\frac{h_2}{AM}}=\frac{h_1}{h_2}=\frac{\frac{1}{2}{h}_{1}AC}{\frac{1}{2}{h}_{2}AF}=\frac{CM}{MF}=\frac{\frac{1}{2}{L}_c}{\frac{1}{2}{L}_c+L_{c}t}=\frac{L_c}{L_c(1+2t)}=\frac{1}{1+2t}\mathrm{˂}\frac{1}{3}$.

Here t > 1 (Figure 7).

Thus, when n >2 cosα ≠$\frac{sin{\alpha }_1}{sin{\alpha }_2}$In a triangle, side CF faces an obtuse angle, i.e. she is the biggest side.

Here СМ = MD =$\frac{1}{2}{L}_c$. DF=DE+EF

Since CD=DE=$L_c$, тo DF= $L_c$+ EF. That is DF >$L_c\Rightarrow t=\frac{DE}{L_c}>1$Thus, we get that, $cos\beta =\frac{a^{n-1}}{c^{n-1}}$ and $cos\alpha =\frac{b^{n-1}}{c^{n-1}}$ only when n=2. However, the cosines$\left(cos\beta ,cos\alpha \right)$ take values close respectively to the values $\frac{a^{n-1}}{c^{n-1}}$ and $\frac{b^{n-1}}{c^{n-1}}$ Let's write the following expressions:

Here $f_1$and $f_2$are functions of positive integers a,b,c and n. Let us determine the values of these functions. To do this, from the triangles ∆ACD and ∆BCD we find cosα and cosβ in the following form:

We substitute in (32) and (33) the value $L_c^2=\frac{a^2{b}^n+a^n{b}^2}{c^n}-\frac{a^n{b}^2}{c^{2n-2}},$ derived from (26)

we get: $cos\beta =\frac{a^{n-1}}{c^{n-1}}{f}_1$ and $cos\alpha =\frac{b^{n-1}}{c^{n-1}}{f}_2$ where

As seen, for n=2 we have:

Then $cos\beta =\frac{a^{n-1}}{c^{n-1}}{f}_1=\frac{a}{c},cos\alpha =\frac{b^{n-1}}{c^{n-1}}{f}_2=\frac{b}{c}.$That is, for a right triangle, everything is correct. Now suppose that n>2, nϵN.

In this case, substituting (31) into (15), we obtain the equality

Thus, for any integer triangular numbers a, b, c and natural powers of n, we can find such functions $f_1$ and $f_2$, which, when formulated, give equality (36).

If we pay attention to the functions (34) and (35), we see that for natural values of a, b, c and n, the functions f_1 and f_2 must be rational. This can also be seen from (36).

Now suppose that there are values n(n>2) under which the condition $a^n+b^n=c^n,$ is satisfied, that is, such values $f_1$ and $f_2 \left(f_1\ne 1, f_2\ne 1\right),$, under which the following conditions were simultaneously met:

$\left\{\begin{array}{c}{a}^n+b^n=c^n \\ a^n{f}_1+b^n{f}_2=c^n \end{array}\right.$ Then, subtracting the first from the equation from the second, we get $a^n\left(1-f_1\right)+b^n\left(1-f_2\right)=0$ or $a^n\left(1-f_1\right)=b^n\left(f_2-1\right).$Where we get

Now we can get the following expression:

Formulas (37) and (38) can be written as:

They can also be presented, for example, as:

Let's note:

1) For n=2, that is, when the triangle is right-angled, regardless of a, b, c, we get $f_1=f_2=1.$ Then the expression $a^2+b^2=c^2$ can have positive integer solutions for a, b, c.

2) With a=b, that is, when the triangle is isosceles, regardless of a, b, c, we get $f_1=f_2=1$. However, then we have $a^n+a^n=c^n$ and $2a^n=c^n$, this case is impossible if a and c are rational.

In formula (36), for an arbitrary triangle, an equality is obtained in which a, b, c, n can receive not only integer values, but also any real positive values.

And now consider the case for which the line of n-lines is perpendicular to the base of the triangle.

As can be seen from Figure 5 the projection of “b” on side “c” is $AH=b·cos\alpha$, and the projection of a on side “c” is $BH=a·cos\beta$, where $c=a·cos\beta +b·cos\alpha$

From formula (41) one can also obtain the formula

On the other hand, from formula (34) we have: $({\frac{b}{c})}^{n-1}=\frac{1}{f_2}·cos\alpha ,({\frac{a}{c})}^{n-1}=\frac{1}{f_1}·cos\beta$If we divide the corresponding sides of these formulas into each other, then we get:

or

If we divide the corresponding sides of formulas (43) and (42) into each other, then we obtain:

Now consider the n-line CN. As is known:$AN=\frac{b^n}{c^{n-1}},BN=\frac{a^n}{c^{n-1}}$Formula (34) can be written in the following form: $f_1=\frac{a·cos\beta }{\frac{a^n}{c^{n-1}}},f_2=\frac{b·cos\alpha }{\frac{b^n}{c^{n-1}}}$

Let's find the length of NH:

Then the length of the n-line L_c can be found as follows:

And now, using the obtained formulas, we will try to prove some statements related to Fermat's theorem.

As we have already noted, formula (36) is true for any triangle with any positive real sides a,b,c and real values “n”.

As we have noted, $f_1$ and $f_2$must be rational.

We are only interested in triangles with natural values a, b, c and “n”.

Note also that for positive integer values a,b,c, the expressions cosα, cosβ, cosγ must be rational. This is clear from the cosine theorem.

Consider the case

From formula (45) we get that this is possible only for n=k, i.e. equality must hold

On the other hand, as can be seen from formulas (47) and (48), with $f_1=f_2=1,$ we have Δ=0.

From (49) we obtain that L_c = h_c. That is, by formulas (45), (47), (48) we obtain that

in this case the line of n-straight lines must be perpendicular to the base. Then formula (26) is transformed into the following form

Here we have: