Today's result is based on work presented by Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Today's blog shows the details to Step 1 in Fermat's Last Theorem n = 4. As before, I use Greek letters to refer to Gaussian Integers and Latin letters to refer to rational integers. I introduced the Gaussian Prime λ in a previous blog.

Lemma: if Fermat's Last Theorem is true for n = 4, then there exist α, β, γ, ε, λ such that:

ε * λ

^{4r}* α

^{4}+ β

^{4}= γ

^{2}where r ≥ 2

(1) Assume that x

^{4}+ y

^{4}= z

^{2}where xyz ≠ 0.

(2) Setting α = x + 0*i, β = y + o*i, γ = z+0*i, we get:

α

^{4}+ β

^{4}= γ

^{2}where α, β, γ, are Gaussian Integers.

(3) Now, we can assume that α, β are coprime.

(a) Let δ = gcd(α,β). [See here for method of gcd for Gaussian Integers]

(b) Then, there exists α',β' such that α = δ * α', β = δ * β'.

(c) We know that gcd(α',β') = 1. [since we already divided out all common divisors]

(d) We also know that δ

^{4}must divide γ

^{2}since γ

^{2}= δ

^{4}(α'

^{4}+ β'

^{4})

(e) Which gives us that δ

^{2}must divide γ [From here based on Euclid's Lemma for Gaussian Integers and Fundamental Theorem of Arithmetic for Gaussian Integers]

(f) Then, there exists γ' such that γ = γ' * δ

^{2}.

(g) So that we get α'

^{4}+ β'

^{4}= γ'

^{2}

(h) This shows that we can always reduce α,β, to coprime values.

(4) We can further assume that α, β, γ are coprime.

(a) We know for example that α

^{2},β

^{2}, and γ are coprime. [See here ]

(b) Assume that there exists δ = gcd(α,γ) where δ is greater than 1.

(c) Then δ

^{2}would necessarily divide α

^{4}and γ

^{2}

(d) It would then also divide β

^{2}which goes against (a) so we reject our assumption in (b).

(e) We can use the same argument to show that gcd(β,γ)=1.

(5) We know that λ divides either α or β [See here for proof.]

(6) Let us assume that it divides α. The same arguments will hold if it divides β

(7) Then, there exists α' such that α = α'*λ

^{r}where r ≥ 1.

(8) Let ε be some unit such as 1, -1, i, or -i. We can now state that:

ε * α'

^{4}* λ

^{4*r}+ β

^{4}= γ

^{2}.

(9) Now, since λ does not divide β, we know that β ≡ 1 (mod λ

^{6}) which also means that β ≡ 1 (mod λ

^{4}) since:

(a) β ≡ 1 (mod λ

^{6}) [See here for proof]

(b) β ≡ 1 (mod λ

^{6}) implies that λ

^{6}divides β - 1. [Definition of ≡]

(c) Thus, there exists δ such that β - 1 = λ

^{6}* δ = λ

^{4}(λ

^{2}* δ).

(10) From (8) and (9), we conclude that:

γ

^{2}≡ 1 (mod λ

^{4}).

(11) Now, based on this we can conclude that γ ≡ 1 (mod λ

^{2}) or γ ≡ i (mod λ

^{2}) since:

(a) λ

^{2}= -2i. [Since λ = 1-i]

(b) γ = a + bi which gives us 4 cases that we need to prove.

(c) Case 1: a is even, b is even. This is impossible since 2 would then divide γ and 2 = (-2i)*i.

(d) Case 2: a is odd, b is odd. This is also impossible since then γ ≡ λ (mod λ

^{2}) and we know that λ does not divide γ

(e) Case 3: a is odd, b is even. Since b is divisible by 2i and a is partially divisible by 2, this gives us: γ ≡ 1 (mod λ

^{2})

(f) Case 4: a is even, b is odd. Since a is divisible by 2i and b is partially divisible by 2i, this gives us: γ ≡ i (mod λ

^{2})

(12) But, γ ≡ i (mod λ

^{2}) is impossible since:

(a) λ

^{2}would divide γ - i.

(b) Then, there exists δ such that: γ - i = λ

^{2}* δ

(c) We can restate this as: γ = λ

^{2}* δ + i.

(d) Squaring both sides gives us:

γ

^{2}= λ

^{4}*δ

^{2}+ 2*λ

^{2}*δ*i - 1 = λ

^{4}*δ

^{2}+ (i*λ

^{2})*λ

^{2}*δ*i - 1 =

λ

^{4}(δ

^{2}- δ) - 1.

(e) But then γ

^{2}+ 1 is divisible by (λ

^{4}) which contradicts step #10 since γ

^{2}- 1 is divisible by λ

^{4}.

(13) So, γ ≡ 1 (mod λ

^{2}) and there exists δ such that:

γ - 1 = λ

^{2}* δ

(14) Adding

**2**to both sides gives us:

γ + 1 = λ

^{2}* δ + 2 = λ

^{2}(δ + i).

(15) Now we can conclude that λ divides δ(δ + i) since

(a) if λ divides δ, then this is true so we can assume that λ does not divide δ

(b) By the definition of Gaussian Integers, there exists a,b such that δ = a + bi.

(c) We know that a,b cannot both be even. If both are even, the δ is divisible by 2 which is divisible by λ

^{2}

(d) We know that a,b cannot both be odd. If both are odd, then we get a remainder of 1 + i which is equal to (λ)*i = (1 - i)*i = i + 1.

(d) If a is even, b is odd, the remainder is i, then λ divides i + i = 2i which is divisible by λ

^{2}.

(e) If a is odd, b is even, the remainder is 1. then λ divides 1 + i which is equal to λ * i.

(16) Multiplying Step#13 to both sides of Step#14 gives us:

γ

^{2}- 1 = λ

^{4}(δ)(δ + i).

(17) From this, we can conclude that λ

^{5}divides γ

^{2}- 1. [From Step #15]

(18) And this gives us that λ

^{5}divides ε * α'

^{4}* λ

^{4*r}+ (β

^{4}-1).

(19) λ

^{5}divides β

^{4}-1 since:

(a) λ does not divide β [from step #5 and step #6 above]

(b) Then, β

^{4}≡ 1 (mod λ

^{6}) [See Lemma 5, here]

(c) So, λ

^{6}divides β

^{4}- 1

(d) Which means that λ

^{5}must also divide β

^{4}- 1

(20) So, λ

^{5}divides ε * α'

^{4}* λ

^{4*r}. [if any x divides A+B (step #18 above) and x divides B (step #19 above), then x divides A+B-B=A]

(21) But λ does not divide α' (step #7 above) and it does not divide ε (see step #8 above), so we are left with the conclusion that λ

^{5}must divide λ

^{4*r}. [Note, ε is a unit and is only divisible by other units]

(22) And this shows that r ≥ 2.

QED

## 4 comments:

Step (18) is bit tricky.

No explaination given of why λ^5 divides(β^4-1).

Think I figured it:

λ doesn't divide β

So λ doesn't divide β^4

So β^4 ≡ 1 (mod λ^6)

So β^4 - 1 = λ^6*ζ (For some ζ)

So β^4 - 1 = λ^5*(λ*ζ)

So λ^5 divides(β^4-1)

Rob

Sorry, step (19)

Hi Rob,

Thanks for your comment. I've modified the blog to make the argument clearer and to provide the explanation that you are looking for.

Cheers,

-Larry

Enjoying this immensely.

In Step (7) you don't explicitly state that

α'andλare coprime, which is used in step (21).Rob

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