# On Fermat's Last Theorem

Carlos Villacres— carlosvillacresjr@outlook.com

May 30, 2024

## Abstract

Here we approach the problem of FLT using the Binomial Theorem and two cases: n even or odd. In this paper we will show that the polynomial $(a+b-c)^n$ can be factored iff $a^n + b^n = c^n$. Using this factored form we attain a general solution for FLT.

# 1 Fermat's Last and the Binomial Theorem

$$1 \quad a, b, c \in \mathbb{R}^+ \text{ and}$$

$$2 \quad n \ge 2 \in \mathbb{Z}^+$$

$$3 \quad (a+b-c)^n = \sum_{j=0}^{n} \binom{n}{j} (-c)^j (a+b)^{n-j}$$

## 1.1 n, even

Suppose n is even, we get that:

$$6 \quad = c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-c)^j (a+b)^{n-j} + (a+b)^n$$

Now we expand the last term,

$$8 \quad (a+b)^n = a^n + \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j} + b^n$$

So,

$$10 \quad (a+b-c)^n = c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-c)^j (a+b)^{n-j} + a^n + \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j} + b^n,$$

$$11 \quad a^n + b^n = c^n \Rightarrow$$
$$ \begin{aligned} (a+b-c)^n &= 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-c)^j (a+b)^{n-j} + \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j} \\ &= 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}] \end{aligned} \quad (1) $$

13 If we can show that this polynomial is divisible by $(c-a)$, then it must also be divisible by $(c-b)$ since a and b are interchangeable. To do this, we will look at the same polynomial, but expanded differently.

$$ (a+b-c)^n = (-1)^n (c-a-b)^n = (c-a-b)^n \implies $$

$$ \begin{aligned} & 16 = b^n + \sum_{j=1}^{n-1} \binom{n}{j} (-b)^j (c-a)^{n-j} + a^n + \sum_{j=1}^{n-1} \binom{n}{j} (-a)^j (c)^{n-j} + c^n \end{aligned} $$

$$ \begin{aligned} & 17 = 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-b)^j (c-a)^{n-j} + (-a)^j c^{n-j}] \end{aligned} $$

19 This shows that if $(c-a)$ is a factor of the polynomial, we only need to look at the second part of the sum along with the leading term to check.

21 We must show that

$$ (c-a) | 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-a)^j c^{n-j}. $$

23 If we plug in $c = a$ and get this equal to 0, then the original polynomial has a factor of $(c-a)$ (as well as $(c-b)$) for all $n$.

25 We get that $c = a \implies$

$$ 2a^n + \sum_{j=1}^{n-1} \binom{n}{j} (-a)^j a^{n-j} = 2a^n + \sum_{j=1}^{n-1} \binom{n}{j} (-1)^j a^j a^{n-j} = 2a^n + a^n \sum_{j=1}^{n-1} \binom{n}{j} (-1)^j $$

28 If we look at Pascal's Triangle, we can clearly see why this alternating sum would be $-2$. Let's look at the 5th and 6th row of Pascal's Triangle as an example when $n = 6$.

31 For $n = 6$, the terms of the polynomial would be

$$ 32 2a^6 + a^6(-6 + 15 - 20 + 15 - 6). $$

33 This can be rewritten with the 5th line of Pascal's Coefficients:
$$34 \quad 2a^6 + a^6(-(1+5) + (5+10) - (10+10) + (10+5) - (5+1)).$$

35 So we can see that no matter what even $n^{th}$ row we are in (without the 1's)

36 we can use the $n-1^{th}$ row to rewrite the sum and show all middle coefficients

37 cancel except the leading and last 1, so we get that

$$38 \quad \sum_{j=1}^{n-1} \binom{n}{j} (-1)^j = -2 \text{ for all even n.}$$

$$39 \quad \text{This} \implies 2a^n + a^n \sum_{j=1}^{n-1} \binom{n}{j} (-1)^j = 0 \text{ for all n, even.}$$

40 This shows us that $(c-a)$ and $(c-b)$ are factors of the original equation. Finally,

41 we get that for n, even:

$$42 \quad (a+b-c)^n = (c-a)(c-b)g_1(n) \text{ where}$$

$$43 \quad g_1(n) = \frac{2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}]}{(c-a)(c-b)}.$$

44 We note here that $c-a$ and $c-b$ divide this polynomial just once each for any

45 $n$. In other words, $g_1$ is not a rational equation and each term has an integer

46 coefficient.

## 1.2 n, odd

48 For $n$ odd, we do something similar. We get that

$$49 \quad (a+b-c)^n = -c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-c)^j (a+b)^{n-j} + a^n + \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j} + b^n$$

50 And $a^n + b^n = c^n \implies$

$$51 \quad (a+b-c)^n = \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}]$$
$$ = (a+b) \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j-1} + \frac{a^j b^{n-j}}{(a+b)}] \quad (2) $$

52 We can show that $(a+b) \mid \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j}$ by plugging in $a=-b$. If the result is zero, then $(a+b)$ is a factor.

$$ 54 \quad \sum_{j=1}^{n-1} \binom{n}{j} (-b)^j b^{n-j} = b^n \sum_{j=1}^{n-1} \binom{n}{j} (-1)^j = b^n 0 = 0 $$

55 This is, again, because the odd rows of Pascal's Triangle would cancel each other out as each term would have its negative in the same row.

Let's define $g(n)$ s.t.

$$ g(n) = \begin{cases} (c-a)(c-b)g_1(n), & \text{if } n \text{ is even} \\ (a+b)g_2(n), & \text{if } n \text{ is odd.} \end{cases} $$

57 Where $g_1(n) =$

$$ 58 \quad \frac{2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}]}{(c-a)(c-b)}, $$

59 and $g_2(n)=$

$$ 60 \quad \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j-1} + \frac{a^j b^{n-j}}{(a+b)}]. $$

## 1.3 Fermat's Last Theorem, proof

61 We have that $(a+b-c)^n = (c-a)(c-b)g_1(n)$ for $n$, even. And $(a+b-c)^n = (a+b)g_2(n)$ for $n$, odd.

62 $n$, even

63 This means that if $a, b, c$ are integers, then $k = a+b-c = [(c-a)(c-b)g_1(n)]^{1/n}$, $\implies$ that $k$ has to be a multiple of the integer $g_1(n)$. So we get that,

$$ 67 \quad k = g_1(n) \left( \frac{(c-a)(c-b)}{g_1(n)^{n-1}} \right)^{1/n} . $$
We'll show this only works for $n = 2$, where $g_1(2) = 2$.

$$69 \quad k = 2\sqrt{\frac{(c-a)(c-b)}{2}}$$

We can let

$$71 \quad a = (c-b) + g(2)^{1/2},$$

$$72 \quad b = (c-a) + g(2)^{1/2},$$

$$73 \quad c = (a+b) - g(2)^{1/2}$$

and define r,s such that

$$75 \quad r = (c-a)^{1/2}, s = [2(c-b)]^{1/2}.$$

So we get

$$77 \quad a = s^2/2 + rs$$

$$78 \quad b = r^2 + rs$$

$$79 \quad c = s^2/2 + r^2 + rs$$

Finally we get $k = 2\sqrt{\frac{(c-a)(c-b)}{2}} = 2(rs/2) = rs$ with the condition that $rs$ is an even integer.

For $n = 4$ we get that $g_1(4) = 2(c-a)(c-b) + 4(a^2 + ab + b^2) \Rightarrow$

$$83 \quad k = g_1(4)\sqrt[4]{\frac{(c-a)(c-b)}{[2(c-a)(c-b) + 4(a^2 + ab + b^2)]^3}}.$$

Upon close inspection, that radical will always be less than 1. This means that for $n \ge 4$, a+b-c is not a factor of the integer $g_1(n)$. So we get a contradiction that a,b,c were all integers.

n, Odd

For $n = 3$ we get that $g(3) = 3(a+b)(c-a)(c-b)$

$$89 \quad a = (c-b) + g(3)^{1/3}, b = (c-a) + g(3)^{1/3}, \text{ and } c = (a+b) - g(3)^{1/3}.$$
We will define $r, s, t$ so that $r = (c-a)^{1/3}, s = (c-b)^{1/3}$, and $t = g(3)^{1/3}/rs = [3(a+b)]^{1/3}$. This means that we get

$$ (a+b-c)^3 = 3(a+b)(c-a)(c-b) \text{ or} $$

$$ (rst)^3 = 3(r^3 + s^3 + 2rst)(r^3)(s^3) \implies $$

$$ t^3 = 3(r^3 + s^3 + 2rst) \implies \text{the cubic} $$

$$ t^3 - 6rst - 3(r^3 + s^3) = 0. $$

Let $p$ be a factor of $r$ or $s$ WLOG. Using Eisenstein's criterion with that prime number, we can see this cubic is irreducible. $t$ has no integer solutions.

This cubic is interesting because of its form similar with the family of elliptic curves $y^2 = x^3 + ax + b$. In any case, the way we show $n=5$ also applies for $n=3$, but the elliptic equation seemed worth noting.

For $n = 5, g_2(5) = 5(c-a)^2(c-b)^2 + 5(a^2 + b^2)(c-a)(c-b)$ which would imply that

$$ k = a + b - c = g_2(5) \sqrt[5]{\frac{a+b}{[5(c-a)^2(c-b)^2 + 5(a^2 + b^2)(c-a)(c-b)]^4}} $$

Again, we can see that the radical is less than 1, so since $g_2(5)$ is not a multiple of $a+b-c$, then $a+b-c$ can't be integers. This applies for $n \ge 3$.

End proof.

Note: We could also show that for $n$ odd, $g(n)$ is also factorable by $(c-a)(c-b)$ for all $n$ odd (and thus all $n$). This would generalize the proof further. However, for $n$ odd, given that it was divisible by $(a+b)$ was easier to show and enough.

## 2 n=2

$$ (a+b-c)^2 = g(2) = 2(c-a)(c-b) \quad (3) $$
Because of the relevance of right triangles, we get trigonometry.

$$a = k(\cos\theta),$$

$$b = k(\sin\theta),$$

$$c = k$$

⇒

$$(\cos\theta + \sin\theta - 1)^2 = 2(1 - \cos\theta)(1 - \sin\theta)$$

$$y = (\cos\theta + \sin\theta - 1)^2 = 2(1 - \cos\theta)(1 - \sin\theta)$$

<figure><img src="image_1.png" /><figcaption>Figure 1: This shows the identity as a function of theta. Notice the identity is $\ge 0$. It also has an interesting rhythm to it.</figcaption></figure>
<figure><img src="image_2.png" /><figcaption>Figure 2: The derivative resembles the rhythm of a heartbeat.</figcaption></figure>
## 2.1 General form of the $n^{th}$ derivative of the trig identity

Here we will simply look at the form of the $n^{th}$ derivative since it has a consistent form.

$$115 \qquad y = (\cos\theta + \sin\theta - 1)^2$$

Let $y^{(n)}$ represent the $n^{th}$ derivative. Then,

$$y^{(n)} = \begin{cases} 2(\cos\theta + \sin\theta - 2^n \sin\theta \cos\theta)(-1)^{\frac{n}{2}-1}, & \text{if } n \text{ is even} \\ (2^n(\cos\theta + \sin\theta) - 2)(\cos\theta - \sin\theta)(-1)^{\frac{n-1}{2}}, & \text{if } n \text{ is odd.} \end{cases}$$

When we investigate the zeroes of the derivative forms, we obtain some interesting trig problem:

For n, even, setting $y^{(n)} = 0$ gives us that

$$120 \qquad \cos\theta + \sin\theta = 2^n \cos\theta \sin\theta.$$

Solving this using trig substitutions to change everything in terms of $\sin(\theta+\pi/4)$ gives us a general solution for the zeros.

$$123 \qquad \theta = \sin^{-1}[2^{-(n+\frac{1}{2})} \pm (2^{-1} + 2^{-1-2n})^{\frac{1}{2}}] - \pi/4 + 2\pi k \text{ and}$$

$$124 \qquad \theta = 3\pi/4 - \sin^{-1}[2^{-(n+\frac{1}{2})} \pm (2^{-1} + 2^{-1-2n})^{\frac{1}{2}}] + 2\pi k \text{ for any integer, } k.$$

For n, odd we get an easier trig problem finding the zeros of

$$126 \qquad \cos\theta + \sin\theta = 2^{1-n} \text{ and } \cos\theta = \sin\theta.$$

## 2.2 Pythagorean Triples and $\sqrt{2}$

$$127 \qquad (a+b-c)^2 = g(2) = 2(c-a)(c-b) \implies$$

We have the Pythagorean Triple generator where s is any even integer, r any integer using the substitution from before:

$$131 \qquad a = \frac{s^2}{2} + rs$$

$$132 \qquad b = r^2 + rs$$

$$133 \qquad c = \frac{s^2}{2} + r^2 + rs$$
$$ \sqrt{2} = \frac{a+b-c}{\sqrt{(c-a)(c-b)}} \quad \text{iff } a^2 + b^2 = c^2 $$

$$ \sqrt[3]{3} = \frac{(a+b-c)}{\sqrt[3]{(a+b)(c-a)(c-b)}} \quad \text{iff } a^3 + b^3 = c^3 $$

### 3 g(n) and the K-Form

<footnote n="134">134 Note: It's actually only necessary that $rs$ is an integer. For example $r = s = \sqrt{2}$.</footnote>

<footnote n="135">135 A special case if $r = s$:</footnote>

<footnote n="136">136 This gives us,</footnote>

<footnote n="137">137 $a = 3\frac{s^2}{2}$</footnote>

<footnote n="138">138 $b = 4\frac{s^2}{2}$</footnote>

<footnote n="139">139 $c = 5\frac{s^2}{2}$</footnote>

<footnote n="140">140 Which is the famous 3,4,5 triple and its multiples.</footnote>

<footnote n="141">141 We can see this when we let $s = \sqrt{2}k_1$ where $k_1 = (c-b)$.</footnote>

<footnote n="142">142 Finally, we also get a form of $\sqrt{2}$ and a form of $\sqrt[3]{3}$.</footnote>

<footnote n="143">143 Which could also be written in an infinite power form since $2 = \frac{(a+b-c)^2}{(c-a)(c-b)}$ and</footnote>

<footnote n="144">144 $2^{-1} = \frac{(c-a)(c-b)}{(a+b-c)^2}$</footnote>

<footnote n="145">145 Let $A = a + b - c$ and $B = (c - a)(c - b)$</footnote>

<footnote n="146">146 $\sqrt{2} = \frac{A}{B2^{-1}} = \frac{A}{B^2} = \dots$</footnote>

<footnote n="148">148 We will end this with a general form of g(n).</footnote>

<footnote n="149">149 We will let $k = (c-a)(c-b)$. Then, we can re-express $g(n)$ as a summation in terms of $k$. First we will look at the case where $n$ is even and then when $n$ is odd. The motivation behind this form is to focus on the reconstruction of the Binomial Coefficients for the $n$'th row we are in. If for $k^j = [(c-a)(c-b)]^j$ we focus on the constant term $(ab)^j$, then it comes clear our choice for $f_j$.</footnote>

<footnote n="154">154 When $n$ is even,</footnote>
$$155 \qquad g(n) = \sum_{j=1}^{n/2} f_j k^j$$

$$156 \qquad f_{n/2} = 2$$

$$157 \qquad f_j = \eta_j \sum_{\xi=0}^{n-2j} a^{\xi} b^{n-2j-\xi} \quad \text{where} \quad \eta_j = \begin{cases} \begin{pmatrix} n \\ j \end{pmatrix}, & \text{if } j=1 \\ \begin{pmatrix} n \\ j \end{pmatrix} - \begin{pmatrix} n \\ j-1 \end{pmatrix}, & \text{else.} \end{cases}$$

And lastly, we will show that for $n \ge 6$ we will also need

$$159 \qquad f_{n/2-1} = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix}$$

$$160 \qquad \text{Where } T_1 = T_3 = \begin{pmatrix} n \\ \frac{n}{2} - 1 \end{pmatrix} - \begin{pmatrix} n \\ \frac{n}{2} - 2 \end{pmatrix} \text{ and } T_2 = \begin{pmatrix} n \\ \frac{n}{2} \end{pmatrix} - \begin{pmatrix} n \\ \frac{n}{2} - 2 \end{pmatrix} - 2.$$

For convenience, we can also express $k^j$ in its convoluted form. Namely,

$$163 \qquad k^j = (a_j * b_j) \cdot c_{2j}$$

$$164 \qquad \text{where } a_j = \left[ \begin{pmatrix} j \\ k_1 \end{pmatrix} (-a)^{k_1} \right], \text{ and } b_j = \left[ \begin{pmatrix} j \\ k_1 \end{pmatrix} (-b)^{j-k_1} \right] \text{ for } k_1 = 0, 1, \dots, j$$

$$165 \qquad c_{2j} = [c^{k_2}] \text{ for } k_2 = 0, 1, \dots, 2j.$$

Let's do some examples. For n=2:

$$167 \qquad g(2) = f_1 k^1 \text{ where } f_1 = 2 \text{ so,}$$

$$168 \qquad g(2) = 2k = 2(c-a)(c-b).$$

For n=4:

$$170 \qquad g(4) = f_2 k^2 + f_1 k^1 \text{ where } f_2 = 2 \text{ and } f_1 = \eta_1 \sum_{\xi=0}^{2} a^{\xi} b^{2-\xi} = 4(a^2 + ab + b^2).$$

$$171 \qquad g(4) = 2k^2 + 4(a^2 + ab + b^2)k$$
172 For $n=6$ (and on) we start using the "trident" $[T_1T_2T_3]$.

$$173 \quad g(6) = f_3 k^3 + f_2 k^2 + f_1 k^1$$

$$174 \quad f_3 = 2 \text{ and } f_2 = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix} \quad \text{where } T_1 = T_3 = 9, T_2 = 12$$

$$175 \quad \text{and } f_1 = \eta_1 \sum_{\xi=0}^{4} a^{\xi} b^{4-\xi} = 6(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$$

$$176 \quad \text{So, } g(6) = 2k^3 + (9a^2 + 12ab + 9b^2)k^2 + 6(a^4 + a^3b + a^2b^2 + ab^3 + b^4)k$$

177 Let's do one more, $n=8$:

$$178 \quad g(8) = f_4 k^4 + f_3 k^3 + f_2 k^2 + f_1 k^1.$$

$$179 \quad f_4 = 2, f_3 = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix} \quad \text{where } T_1 = T_3 = 56-28=28, T_2 = 70-28-2=40$$

$$180 \quad f_2 = \eta_2 \sum_{\xi=0}^{4} a^{\xi} b^{4-\xi} \quad \text{where } \eta_2 = 28-8=20.$$

$$181 \quad \text{So we get } f_2 = 20(a^4 + a^3b + a^2b^2 + ab^3 + b^4).$$

$$182 \quad f_1 = \eta_1 \sum_{\xi=0}^{6} a^{\xi} b^{6-\xi} \quad \text{and } \eta_1 = 8.$$

183 Finally, we get that $g(8) =$

$$184 \quad 2k^4 + (28a^2 + 40ab + 28b^2)k^3 + 20(a^4 + a^3b + a^2b^2 + ab^3 + b^4)k^2 + 8(a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6)k.$$

185 When $n$ is odd,

$$186 \quad \frac{g(n)}{a+b} = \sum_{j=1}^{(n-1)/2} f_j k^j$$

$$187 \quad f_{(n-1)/2} = n$$
$$190 \quad f_j = \eta_j \sum_{\xi=0}^{(n-1-2j)/2} [a^\xi b^{(n-1-2j)/2-\xi}]^2 \text{ where } \eta_j = \begin{cases} \begin{pmatrix} n \\ j \end{pmatrix}, & \text{if } j=1 \\ \begin{pmatrix} n \\ j \end{pmatrix} - \begin{pmatrix} n \\ j-1 \end{pmatrix}, & \text{else.} \end{cases}$$

191 And lastly, we will show that for $n \ge 7$ we will also need

$$192 \quad f_{(n-3)/2} = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix}$$

$$193 \quad \text{Where } T_1 = T_3 = \left( \frac{n}{\frac{n-3}{2}} \right) - \left( \frac{n}{\frac{n-5}{2}} \right) \text{ and } T_2 = \left( \frac{n}{\frac{n-1}{2}} \right) - \left( \frac{n}{\frac{n-3}{2}} \right) - n.$$

194 And some examples for n, odd. When n=3:

$$195 \quad \frac{g(3)}{a+b} = f_1 k^1 \text{ where } f_1 = 3 \text{ so we get}$$

$$196 \quad \frac{g(3)}{a+b} = 3k$$

197 For n=5:

$$198 \quad \frac{g(5)}{a+b} = f_2 k^2 + f_1 k^1 \text{ where } f_2 = 5 \text{ and}$$

$$199 \quad f_1 = \eta_1 \sum_{\xi=0}^{1} [a^{\xi} b^{1-\xi}]^2 = 5(a^2 + b^2) \text{ so we get that}$$

$$200 \quad \frac{g(5)}{a+b} = 5k^2 + 5(a^2 + b^2)k$$

201 For n=7 we start using the 'trident' like before:

$$202 \quad \frac{g(7)}{a+b} = f_3 k^3 + f_2 k^2 + f_1 k^1 \text{ where } f_3 = 7 \text{ and}$$

$$203 \quad f_2 = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix} \text{ where}$$

$$204 \quad T_1 = T_3 = 21 - 7 = 14 \text{ and } T_2 = 35 - 21 - 7 = 7, \text{ and lastly,}$$

$$205 \quad f_1 = \eta_1 \sum_{\xi=0}^{2} [a^{\xi} b^{2-\xi}]^2 = 7(a^4 + a^2 b^2 + b^4). \text{ So,}$$

$$206 \quad \frac{g(7)}{a+b} = 7k^3 + (14a^2 + 7ab + 14b^2)k^2 + 7(a^4 + a^2 b^2 + b^4)k.$$
207 Let's do n=9:

$$
208 \quad \frac{g(9)}{a+b} = f_4 k^4 + f_3 k^3 + f_2 k^2 + f_1 k^1 \text{ where } f_4 = 9 \text{ and}
$$

$$
209 \quad f_3 = [T_1 T_2 T_3] \cdot \begin{bmatrix} a^2 \\ ab \\ b^2 \end{bmatrix} \quad \text{where } T_1 = T_3 = 84 - 36 = 48 \text{ and } T_2 = 126 - 84 - 9 = 33
$$

$$
210 \quad f_2 = \eta_2 \sum_{\xi=0}^{2} [a^{\xi} b^{2-\xi}]^2 = (36-9)(a^4 + a^2 b^2 + b^4)
$$

$$
211 \quad f_1 = \eta_1 \sum_{\xi=0}^{3} [a^{\xi} b^{3-\xi}]^2 = 9(a^6 + a^4 b^2 + a^2 b^4 + b^6). \text{ So we get that } g(9) \text{ is:}
$$

$$
212 \quad \frac{g(9)}{a+b} = 9k^4 + (48a^2 + 33ab + 48b^2)k^3 + 27(a^4 + a^2b^2 + b^4)k^2 + 9(a^6 + a^4b^2 + a^2b^4 + b^6)k
$$

213 References

214 None

215 2020 Mathematics Subject Classification: 11D41