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# On Fermat's Last and the Binomial Theorem

Carlos Villacres

► To cite this version:

Carlos Villacres. On Fermat's Last and the Binomial Theorem. Épijournal de Géométrie Algébrique, 2024, 10.36227/techrxiv.170629872.29614231/v2. hal-04599794

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https://hal.science/hal-04599794

Submitted on 4 Jun 2024

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# On Fermat's Last and the Binomial Theorem

Carlos Villacres

► To cite this version:

Carlos Villacres. On Fermat's Last and the Binomial Theorem. 2024. [hal-04599794](https://hal.science/hal-04599794v1)

HAL Id: hal-04599794

https://hal.science/hal-04599794v1

Preprint submitted on 4 Jun 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

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# 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. $a, b, c \in \mathbb{R}^+$ and

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

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

## 1.1 n, even

Suppose n is even, we get that:

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

7. Now we expand the last term,

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

9. So,

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

11. $a^n + b^n = c^n \Rightarrow$
# 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. $a, b, c \in \mathbb{R}^+$ and

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

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

## 1.1 n, even

Suppose n is even, we get that:

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

7. Now we expand the last term,

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

9. So,

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

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

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.

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

$$ \begin{aligned} & 17 \quad b^n + \sum_{j=1}^{n-1} \binom{n}{j} (-b)^j (c-a)^{n-j} + a^n + \sum_{j=1}^{n-1} \binom{n}{j} (-a)^j (c)^{n-j} + c^n \\ & 18 \quad = 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-b)^j (c-a)^{n-j} + (-a)^j c^{n-j}] \end{aligned} $$

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.

20 We must show that

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

21 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 $n$.

22 We get that $c = a \implies$

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

23 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 $n=6$.

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

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

25 This can be rewritten with the 5th line of Pascal's Coefficients:
$$ \begin{aligned} (a+b-c)^n &= 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} (-c)^j (a+b)^{n-j} + \sum_{j=1}^{n-1} \binom{n}{j} a^j b^{n-j} \\ &= 2c^n + \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}] \end{aligned} \quad (1) $$

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.

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

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

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

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

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

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

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

$$51 \quad (a+b-c)^n = \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j} + a^j b^{n-j}]$$
$$34 \quad 2a^6 + a^6(-(1+5) + (5+10) - (10+10) + (10+5) - (5+1)).$$

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,

$$ 67 \quad k = g_1(n) \left( \frac{(c-a)(c-b)}{g_1(n)^{n-1}} \right)^{1/n} . $$
$$ = (a+b) \sum_{j=1}^{n-1} \binom{n}{j} [(-c)^j (a+b)^{n-j-1} + \frac{a^j b^{n-j}}{(a+b)}] \quad (2) $$

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'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 $n=2$

$$ (a+b-c)^2 = g(2) = 2(c-a)(c-b) \quad (3) $$
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_2.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>
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_2.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_3.png" /><figcaption>Figure 2: The derivative resembles the rhythm of a heartbeat.</figcaption></figure>
<figure><img src="image_3.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.

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

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

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

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

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

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

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

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

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

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

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

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

$$b = r^2 + rs$$

$$c = \frac{s^2}{2} + r^2 + rs$$
## 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.

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

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

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

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

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

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

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

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

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

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

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

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

$$b = r^2 + rs$$

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

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

## 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 $\sqrt{2} = \frac{A}{B^2-1} = \frac{A}{B^2} = \dots$</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</footnote>

<footnote n="150">150 terms of k. First we will look at the case where n is even and then when n is</footnote>

<footnote n="151">151 odd. The motivation behind this form is to focus on the reconstruction of the</footnote>

<footnote n="152">152 Binomial Coefficients for the n'th row we are in. If for $k^j = [(c-a)(c-b)]^j$ we</footnote>

<footnote n="153">153 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>
$$ \sqrt{2} = \frac{a+b-c}{\sqrt{(c-a)(c-b)}} \quad \text{iff} \quad a^2 + b^2 = c^2 $$

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

## 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 $\sqrt{2} = \frac{A}{B^2-1} = \frac{A}{B^2} = \dots$</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:

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

$$171 \qquad g(4) = 2k^2 + 4(a^2 + ab + b^2)k$$
$$155 \qquad g(n) = \sum_{j=1}^{n/2} f_j k^j$$

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

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

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

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

$$160 \qquad \text{Where } T_1 = T_3 = \begin{pmatrix} n \\ \frac{n}{2} - 1 \end{pmatrix} - \begin{pmatrix} n \\ \frac{n}{2} - 2 \end{pmatrix} \text{ and } T_2 = \begin{pmatrix} n \\ \frac{n}{2} \end{pmatrix} - \begin{pmatrix} n \\ \frac{n}{2} - 2 \end{pmatrix} - 2. \qquad 161$$

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

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

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

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

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,

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

$$187 \quad f_{(n-1)/2} = n$$
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

$$181 \quad f_2 = \eta_2 \sum_{\xi=0}^{4} a^{\xi} b^{4-\xi} \quad \text{where } \eta_2 = 28-8=20.$$

$$182 \quad \text{So we get } f_2 = 20(a^4 + a^3b + a^2b^2 + ab^3 + b^4).$$

$$183 \quad f_1 = \eta_1 \sum_{\xi=0}^{6} a^{\xi} b^{6-\xi} \quad \text{and } \eta_1 = 8.$$

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

$$185 \quad 2k^4 + (28a^2 + 40ab + 28b^2)k^3 + 20(a^4 + a^3b + a^2b^2 + ab^3 + b^4)k^2 + 8(a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6)k.$$

$$186$$

187 When $n$ is odd,

$$188 \quad \frac{g(n)}{a+b} = \sum_{j=1}^{(n-1)/2} f_j k^j$$

$$189 \quad f_{(n-1)/2} = n$$
$$190 \quad f_j = \eta_j \sum_{\xi=0}^{(n-1-2j)/2} [a^\xi b^{(n-1-2j)/2-\xi}]^2 \text{ where } \eta_j = \begin{cases} \begin{pmatrix} n \\ j \end{pmatrix}, & \text{if } j=1 \\ \begin{pmatrix} n \\ j \end{pmatrix} - \begin{pmatrix} n \\ j-1 \end{pmatrix}, & \text{else.} \end{cases}$$

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:

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

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

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

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

$$206 \quad \frac{g(7)}{a+b} = 7k^3 + (14a^2 + 7ab + 14b^2)k^2 + 7(a^4 + a^2 b^2 + b^4)k.$$
$$190 \quad f_j = \eta_j \sum_{\xi=0}^{(n-1-2j)/2} [a^\xi b^{(n-1-2j)/2-\xi}]^2 \text{ where } \eta_j = \begin{cases} \begin{pmatrix} n \\ j \end{pmatrix}, & \text{if } j=1 \\ \begin{pmatrix} n \\ j \end{pmatrix} - \begin{pmatrix} n \\ j-1 \end{pmatrix}, & \text{else.} \end{cases}$$

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
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