Euclidean Integers

A quadratic integer is considered Euclidean if it can be characterized by a division algorithm. This is the algorithm which states that division of any integer by an nonzero integer results in two unique values: a quotient and a remainder and the norm of the remainder is always less than the norm of the divisor.

In a previous blog, I gave the proof for this with regard to Gaussian Integers and with regard to rational integers. So, using these proofs, I have shown that both Gaussian Integers and rational integers are Euclidean.

Since the greatest common divisors algorithm from Euclid derives from the division algorithm, we can also conclude that all Euclidean integers have a greatest common denominator that is a linear combination of two other integers. It is also straight forward to show that all Euclidean integers are characterized by unique factorization.

This means that one sure path to establishing unique factorization is to show that a quadratic integer is Euclidean. Interestingly, it turns out that there are quadratic integers which are not Euclidan which still possess unique factorization.

Definition of Euclidean Integer: A quadratic integer a + b√d is Euclidean if for all integers α, β where β is nonzero, there exists a unique value δ such that δ = α - η*β and absolute(Norm(δ)) is less than absolute(Norm(β)).

Lemma: Z[√2], Z[√-2], Z[√3] are Euclidean

(1) For purposes of this proof, assume that √d = √2, √-2, or √3.

(2) There exists a,b,e,f such that [from the definition quadratic integers]
α = a + b√d
β = e + f√d

(3) α/β = (a + b√d)/(e + f√d) = (a + b√d)(e - f√d) /(e² - f²d)=
(ae - af√d + be√d - bdf)/(e² - f²d) =
(ae - bdf)/(e² - f²d) + √d(be - af)/(e² - f²d)

(4) Let r = (ae - bdf)/(e² - f²d), s = (be - af)/(e² - f²d) where r,s are rational but not necessarily integer. [For a review of rational numbers and their properties, see here]

(5) So α/β = r + s√d.

(6) We know that there exists m,n which are rational integers such that:
absolute(r - m) ≤ (1/2) and absolute(s - n) ≤ (1/2). [See here for proof]

(7) Let η = m + n√d, let δ = α - β * η where η, δ are quadratic integers of type Z[√d]. [For a review of quadratic integers, see here.]

NOTE: The keypoint point is to prove that absolute(Norm(δ)) is less than absolute(Norm(β)).

(8) Norm(δ) = Norm(α - β * η) = Norm(β[(α / β) - η) =
Norm(β) * Norm([α/β] - η)

(9) This means that the key point is to prove that absolute(Norm([α/β] - η)) is less than 1.

(10) We know that:
Norm([α/β] - η) = Norm([r + s√d] - [m + n√d]) = Norm([r - m] + [s - n]√d) =
([r - m] + [s - n]√d)([r - m] - [s - n]√d) = (r - m)² - d(s - n)²

(11) Applying step(6),

gives us for d=2:
absolute((r - m)² - d(s - n)²) ≤ absolute((1/2)² - 2(1/2)²) =
absolute((1/4) - 2(1/4)) = absolute(-1/4) = 1/4 which is less than 1.

gives us for d=-2:
absolute((r - m)² - d(s - n)²) ≤ absolute((1/2)² + 2(1/2)²) =
absolute((1/4) + 2(1/4)) = absolute(3/4) = 3/4 which is less than 1.

gives us for d=3:
absolute((r - m)² - d(s - n)²) ≤ absolute((1/2)² - 3(1/2)²) =
absolute((1/4) - 3(1/4)) = absolute(-2/4) = 1/2 which is less than 1.

QED

Lemma: Z[(1+√-3)/2], Z[(1+√-7)/2], Z[(1+√-11)/2], Z[(1+√5)/2] are Euclidean

(1) For purposes of this proof, assume that √d = √-3, √-7 ,√-11 , or √5.

(2) There exists a,b,e,f such that
α = a + b√d
β = e + f√d

(3) α/β = (a + b√d)/(e + f√d) = (a + b√d)(e - f√d) /(e² - f²d)=
(ae - af√d + be√d - bdf)/(e² - f²d) =
(ae - bdf)/(e² - f²d) + √d(be - af)/(e² - f²d)

(4) Let r = (ae - bdf)/(e² - f²d), s = (be - af)/(e² - f²d) where r,s are rational but not necessarily integer. [For a review of rational numbers and their properties, see here]

(5) So α/β = r + s√d.

(6) We also know that there exists p which is a rational integer such that:
absolute(s - p/2) ≤ 1/4. [See here for proof]

(7) We also know that there exists o which is a rational integer such that:
absolute(r - p/2 - o) ≤ 1/2. [See here for proof]

(8) let η = r + s[(1 + √d)/2] which is an integer as explained in a previous blog since in all above cases d ≡ 1 (mod 4).

(9) Let δ = β*([r - p/2 - o] + [s - p/2]√d)

(10) Norm(δ) = Norm(β) * Norm([r - p/2 - o] + [s - p/2]√d)

(11) Once again, to finish this proof, we need only to show that abs(Norm([r - p/2 - o] + [s - p/2]√d)) is less than 1.

(12) Norm([r - p/2 - o] + [s - p/2]√d) = ([r - p/2 - o] + [s - p/2]√d)([r - p/2 - o] -[s - p/2]√d)
= [r - p/2 - o]² - d[s - p/2]² ≤ (1/2)² - d(1/4)² = (1/4) - d/16.

QED

In addition, here are more interesting facts:

• There are exactly 21 types of quadratic integers that are Euclidean: a quadratic integer is Euclidean if d = -11, -7, -3, -2, -1, 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37,41, 57, and 73.

• Z[(-1 + √-19)/2] is not Euclidean but still has unique factorization

• Z[√-5] does not have unique factorization.

The content of this blog is based in a large part on Harold M. Stark's An Introduction to Number Theory.

Quadratic Integers: Generalizing Gaussian Integers

Leonhard Euler proposed a proof for Fermat's Last Theorem: n = 3, using integers based on a+b√-3. Carl Friedrich Gauss showed the need for a proof for uniqueness of factorization for a + b√-1. Both of these integer extensions are examples of what we call today quadratic integers.

Quadratic integers are integers that are extended in the form of a + bα where α is an irrational value that is a solution to a quadratic equation. A quadratic equation is any equation that can be represented in the form x² + bx + c = 0 where b,c are rational integers. The important point is that the coefficient for x² needs to be 1 and the other coefficients need to be rational integers.

Applying basic algebra, we know that quadratic equations have the following solution:

Theorem: ax² + bx + c = 0 → x = (-b ± √b² - 4ac)/2a.

(1) Subtracting c from both sides gives us:
ax² + bx = -c

(2) Multiplying both sides by 4a gives us:
4a²x² + 4abx = -4ac.

(3) Adding b² to both sides, gives us:
4a²x² + 4abx + b² = b² - 4ac.

(4) Now, since (2ax + b)*(2ax + b) = 4a²x² + 4abx + b², we get:
(2ax + b)² = b² - 4ac.

(5) Taking the square root of both sides, gives us:
2ax + b = ± √b² - 4ac

(6) Finally, subtracting b and then dividing both sides by 2a gives us our proof.

QED

From this formula, we see that Euler's integer is a solution to:
x² + 3 = 0.

and Gaussian Integers are a solution to:
x² + 1 = 0.

For purposes of notation, a set of quadratic integers is identified using a Z-notation. Z by itself represents the set of standard integers. Z[i] represents the set of Gaussian Integers. Z[√-3] represents the set of Euler's integers.

Now, when it comes to integers of the form Z[√d] where d ≡ 1 (mod 4), it turns out that an integer is of the form: (a + b√d)/2 where a,b are both odd or both even.

To understand the mathematics behind this result, let's start out with a very simple lemma:

Lemma: a² ≡ 1 or 0 (mod 4).

(1) We know that a ≡ 0, 1, 2, or 3 (mod 4).

(2) Now here's what we get for each of these values in the case of a²

(a) 0² ≡ 0 * 0 ≡ 0 (mod 4).
(b) 1² ≡ 1 * 1 ≡ 1 (mod 4).
(c) 2² ≡ 2 * 2 ≡ 0 (mod 4).
(d) 3² ≡ 3 * 3 ≡ 1 (mod 4).

QED

Here is the lemma that establishes this:

Lemma: for a set of quadratic integers Z[√d] , if d ≡ 1 (mod 4), then the form of this quadratic integer is (a + b√d)/2 where both a,b are even or both a,b are odd, otherwise, it is a + b√d

(1) The goal here is to prove that (a + b√d)/2 is an integer, that is, it satisfies an equation of the form x² + bx + c = 0.

(2) Now let's assume that we have a value α which is equal to (e + f√d)/2 where e,f are rational integers and d ≡ 1 (mod 4).

(3) Let x = (e + f√d)/2

(4) Then,
2x - e = f√d

(5) And squaring each side gives us:
4x² - 4ex + e² = f²d

(6) Which amounts to:
4x² - 4ex + e² - f²d = 0

(7) Now if we divide both sides by 4, we get:
x² - ex + (1/4)[e² - f²d] = 0

(8) Now, if both e,f are odd, then: [from the lemma above]
e² - f²d ≡ 1 - 1*1 ≡ 0 (mod 4).

(9) Now, if both e,f are even, then: [from the lemma above]
e² - f²d ≡ 0 - 0 * 1 ≡ 0 (mod 4).

(10) This value then meets our definition of quadratic integer.

QED

Corollary: if d ≡ 1 (mod 4), a number is an integer if it can be written as a + b[(1 + √d)/2], where a,b are rational integers.

(1) a + b(1 + √d)/2 = [(2a + b) + b√d]/2

If b is odd, then (2a + b) is odd. If b is even, then (2a + b) is even. In both cases (2a + b),b are both even or both odd.

(2) If a,b are both even or both odd.

(a + b√d)/2 = (a - b)/2 + b(1 + √d)/2

QED

The result of this is that in the case of d=-3, the set of integers is Z[(-1+√-3)/2] because we are dealing with integers of the form (a + b√-3)/2. I will go into more detail on this when I revisit a proof for n=3 based on Eisenstein integers.

The other interesting point about quadratic integers is that it turns out that not all of them are characterized by unique factorization and most of them are not characterized by a division algorithm.

I will go into more detail about this in my next blog.

The content of this blog is based in a large part on Harold M. Stark's An Introduction to Number Theory.

Proof for n=4 using Gaussian Integers: Step 2

Today's blog continues a proof that was first presented in a previous blog. If you are new to unique factorization, start here. If you are new to Gaussian Integers, start here. To begin this proof, start here.

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 2 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 Gaussian integers: μ₁, μ₂, β, ε, λ such that:

ε * λ^4(r-1) * μ₂⁴ + μ₁⁴ = β².

(1) From the previous lemma, we know that:
ε * λ^4r*α⁴ = γ² - β⁴ = (γ - β²)(γ + β²)

(2) We know that λ² divides both (γ - β²) and (γ + β²) since:

(a) Assume λ² does not divide γ - β²

(b) Then, it must divide γ + β² [From Euclid's Generalized Lemma for Gaussian Integers and (step #1 above)]

(c) But then it necessarily divides γ - β² since
(γ + β²) + (γ - β²) = 2*γ = (λ²)(i * γ) [Since 2= i*λ²]

(d) Having found a contradiction we reject our assumption in step #2a above. We can make the same exact argument if we assume that λ² does not divide γ + β²

(e) So, we are left concluding that λ² necessarily divides both γ - β² and γ + β².

(3) We can also conclude that gcd(γ - β², γ + β²) cannot be greater than 2.

(a) Assume that gcd(γ - β², γ + β²) is greater than 2. Then gcd = λ³ or there exists a prime μ that is greater than 2. [Since we know that λ divides 2, see Lemma 2, here and further, we know that there is no other prime less than 2; see Lemma 4, here]

(b) First, let's assume that there exists a prime μ that is greater than 2.

(c)Then there exists values δ₁, δ₂ such that:
γ - β² = μ * δ₁ and γ + β² = μ * δ₂.

(d) And this means that: μ divides γ since:
2 * γ = (γ - β²) + (γ + β²) = μ(δ₁ + δ₂).

(e) And this also means that μ divides β² since:
2 * β² = (γ + β²) - (γ - β²) = μ(δ₂ - δ₁).

(f) But this is a contradiction since from the lemma cited in step #1 above (see step #4, here), we know that gcd(β²,γ) = 1.

(g) So we reject our assumption in step #3b above and assume that gcd = λ³.

(h) But, then λ³ divides 2 * γ = (γ - β²) + (γ + β²) and λ³ divides 2 * β².

(i) Since λ² is an associate of 2 (see Lemma 2, here), this gives us that λ divides both γ and β².

(j) But this is impossible since we know that β and γ are relatively prime from the lemma cited in step #1 above (see step #4, here)

(k) So we can also reject the assumption in step #3g above and for that matter, the assumption in step #3a above.

(l) So, we can conclude that gcd(γ - β², γ + β²) cannot be greater than 2.

(4) So, we can conclude that λ^4r-2 divides either γ + β² or γ - β² since:

(a) From step #1 above, we know that: λ^r divides (γ - β²)(γ + β²).

(b) From step #2 above, we know that λ² divides both (γ - β²) and (γ + β²).

(c) But, from step #3 above, we know that (λ^4r/λ² = λ^4r-2) can only divide (γ + β²) or (γ - β²) but not both.

(5) So, there exist two values: let's call them η₁ and η₂ such that:
γ ± β² = λ² * η₁
γ ± β² = λ^4r-2 * η₂

NOTE: Please read ± as + or -. So that one of these values is γ + β² and another of these values is γ - β².

(6) We also can conclude that gcd(η₁,η₂) = 1.

(a) Assume that gcd(η₁,η₂) = δ which is not a unit.

(b) So that δ ≥ λ [Since there is no other prime less than λ, see Lemma 4, here]

(c) But then δ*λ² would be greater than 2 which contradicts Step (3) above.

(7) Thus, ε*λ^4rα⁴ = λ^4rη₁η₂ since:

ε*λ^4rα⁴ =(γ - β²)(γ + β²) [From step #1 above]

λ^4rη₁η₂ =(γ - β²)(γ + β²) [From step #5 above]

(8) By the lemma of relatively prime divisors of n-powers, we can conclude that there exists μ₁ and μ₂ such that:

η₁ = ε₁ * μ₁⁴
η₂ = ε₂ * μ₂⁴

NOTE: ε₁ and ε₂ are units.

(9) Now, we know that:
2*β² = γ + β² - (γ - β²)

(10) From step(5) above and step(8) above, let's assume that:
γ - β² = ε₁ * λ² * μ₁⁴
γ + β² = ε₂ * λ^4r-2 * μ₂⁴

NOTE: We can make a similiar argument if the other case is true.

(11) Then,
2 * β² = ε₂ * λ^4r-2 * μ₂⁴ - ε₁ * λ² * μ₁⁴

(12) Which dividing both sides by i*λ² gives us:
β² = -i*ε₂*λ^4r-4*μ₂⁴ + i*ε₁*μ₁⁴

[Note: since 2=i*λ², see Lemma 2, here and since 1/i=-i since -i*i=-(-1)=1]

(13) Now, we can assume that i*ε₁ = 1 since:

(a) r ≥ 2, so we know that λ⁴ divides β² - i*ε₁*μ₁⁴.

(b) But, λ does not divide β

[if it did, from step #2 above, this would mean λ also divides γ, but this is impossible since they are relatively prime from the lemma cited in step #1 above (see step #4, here)]

(c) So, λ cannot divide μ₁

[if λ divided μ₁, then it would also divide η₁ from step #8 above and this is impossible since gcd(η₁,λ)=1 from step #5 above.]

(d) So μ₁ ≡ 1 (mod λ⁶) [See Lemma 5, here]

(e) So that, λ⁴ divides β² - i*ε₁.

(f) But, λ⁶ divides β⁴ - 1 = (β² - 1)(β² + 1)

(g) So, β² ≡ 1 or -1 (mod λ⁴)

(h) This shows that i*ε₁ = ± 1.

(i) If μ₁ = -1, by multiplication with -1, we obtain the relation:
-i*ε₁λ^4(r-1)μ₂⁴ + μ₁⁴ = (iβ)².

(14) Combining Step (12) and Step(13) and setting ε = -i*ε₁ gives us:
β² = ε*λ^4(r-1)*μ₂⁴ + μ₁⁴

QED

Proof for n=4 using Gaussian Integers: Step 1

Today's blog continues a proof that was first presented in a previous blog. If you are new to unique factorization, start here. If you are new to Gaussian Integers, start here. To begin this proof, start here.

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 * α⁴ + β⁴ = γ² where r ≥ 2

(1) Assume that x⁴ + y⁴ = z² where xyz ≠ 0.

(2) Setting α = x + 0*i, β = y + o*i, γ = z+0*i, we get:
α⁴ + β⁴ = γ² 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 δ⁴ must divide γ² since γ² = δ⁴(α'⁴ + β'⁴)
(e) Which gives us that δ² 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' * δ².
(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 α²,β², and γ are coprime. [See here ]
(b) Assume that there exists δ = gcd(α,γ) where δ is greater than 1.
(c) Then δ² would necessarily divide α⁴ and γ²
(d) It would then also divide β² 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 + β⁴ = γ².

(9) Now, since λ does not divide β, we know that β ≡ 1 (mod λ⁶) which also means that β ≡ 1 (mod λ⁴) since:

(a) β ≡ 1 (mod λ⁶) [See here for proof]
(b) β ≡ 1 (mod λ⁶) implies that λ⁶ divides β - 1. [Definition of ≡]
(c) Thus, there exists δ such that β - 1 = λ⁶ * δ = λ⁴(λ² * δ).

(10) From (8) and (9), we conclude that:
γ² ≡ 1 (mod λ⁴).

(11) Now, based on this we can conclude that γ ≡ 1 (mod λ²) or γ ≡ i (mod λ²) since:

(a) λ² = -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 λ²) 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 λ²)
(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 λ²)

(12) But, γ ≡ i (mod λ²) is impossible since:

(a) λ² would divide γ - i.
(b) Then, there exists δ such that: γ - i = λ² * δ
(c) We can restate this as: γ = λ² * δ + i.
(d) Squaring both sides gives us:
γ² = λ⁴*δ² + 2*λ²*δ*i - 1 = λ⁴*δ² + (i*λ²)*λ²*δ*i - 1 =
λ⁴(δ² - δ) - 1.
(e) But then γ² + 1 is divisible by (λ⁴) which contradicts step #10 since γ² - 1 is divisible by λ⁴.

(13) So, γ ≡ 1 (mod λ²) and there exists δ such that:
γ - 1 = λ² * δ

(14) Adding 2 to both sides gives us:
γ + 1 = λ² * δ + 2 = λ²(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 λ²
(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 λ².
(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:
γ² - 1 = λ⁴(δ)(δ + i).

(17) From this, we can conclude that λ⁵ divides γ² - 1. [From Step #15]

(18) And this gives us that λ⁵ divides ε * α'⁴ * λ^4*r + (β⁴-1).

(19) λ⁵ divides β⁴-1 since:

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

(b) Then, β⁴ ≡ 1 (mod λ⁶) [See Lemma 5, here]

(c) So, λ⁶ divides β⁴ - 1

(d) Which means that λ⁵ must also divide β⁴ - 1

(20) So, λ⁵ 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 λ⁵ 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

i - 1 and Fermat's Last Theorem n = 4

Today's blog continues a proof that was first presented in a previous blog. If you are new to unique factorization, start here. If you are new to Gaussian Integers, start here. To begin this proof, start here.

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

Today's blog applies the Gaussian prime λ to 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 α⁴ + β⁴ = γ² where gcd(α,β,γ)=1, then we can assume that λ divides α * β

(1) Assume λ does not divide α * β

(2) Then, from a previous lemma, we know that α⁴ ≡ 1 (mod λ⁶), β⁴ ≡ 1 (mod λ⁶)

(3) Then, there exists δ₁, δ₂ where: [based on the definition of ≡ found here]
α⁴ - 1 = λ⁶ * δ₁.
β⁴ - 1 = λ⁶ * δ₂

(4) So that:
α⁴ = λ⁶*δ₁ + 1.
β⁴ = λ⁶*δ₂ + 1.

(5) Then:
γ² = λ⁶*δ₁ + 1 + λ⁶*δ₂ + 1 = λ⁶(δ₁ + δ₂) + 2.

(6) Since 2 = i*λ², we get: [From a lemma, found here]
γ² = λ⁶(δ₁ + δ₂) + i*λ² =λ²[λ⁴(δ₁ + δ₂) + i]

(7) Which shows that:
λ² divides γ²

(8) And from this, that:
λ divides γ [Since we have a Gaussian Integer proofs for gcd, unique factorization, and Euclid's Lemma, we can use the proof here. ]

(10) This means that λ⁴ does not divide γ².

(a) Assume that λ⁴ does divide γ².
(b) Then λ⁴ would divide 2 since γ² = λ⁴(λ²δ₁ + λ²δ₂) + 2. [Since d divides a + b and d divides a, implies d divides b]
(c) Then there exists ζ such that 2 = λ⁴ * ζ
(d) Since 2 = i *λ², we get:
i *λ²= λ⁴*ζ
(e) Dividing both sides from λ² gives:
i = λ²*ζ = (1-i)²*ζ = -2i * ζ
(f) Which is impossible since ζ is a Gaussian Integer and not a fraction.

(11) This also gives us that λ² does not divide γ [if it did, then λ⁴ would divide γ² which it cannot.]

(12) Now, since λ divides γ, there must exist a value η such that γ = λ * η

(13) We know that λ cannot divide η. [If it did, then λ² would divide γ]

(14) Now, we have our contradiction. Here are the details

(15) In step(5), we showed that:
γ² = λ⁶*δ₁ + 1 + λ⁶*δ₂ + 1 = λ⁶(δ₁ + δ₂) + 2.

(16) Letting μ = δ₁ + δ₂ and letting γ = λ * η, we get:
(λ*η)² = λ²*η² = λ⁶ * μ + i*(λ²)

(17) Dividing both sides by λ², we get:
η² = λ⁴ * μ + i.

(18) Squaring both sides gives us:
η⁴ = (λ⁴ * μ + i)² = λ⁸*μ² + 2*λ⁴*μ*i -1 =
λ⁸*μ² + ( i*λ²)*λ⁴*μ*i -1 =
λ⁶[λ²*μ² -μ] - 1

(19) So, that we get:
η⁴ ≡ -1 (mod λ⁶)

(20) But, since λ does not divide η, from a previous lemma, we get:
η⁴ ≡ 1 (mod λ ⁶)

(21) It is impossible for both of these modulo λ⁶ values to be true.

(a) Assume η⁴ ≡ -1 (mod λ⁶) and η⁴ ≡ 1 (mod λ⁶).
(b) λ⁶ divides both η⁴ - 1 and η⁴ + 1.
(c) Then there exists ν₁ and ν₂ such that;
η - 1 = λ⁶ * ν₁
η + 1 = λ⁶ * ν₂
(d) Adding the two equations together gives us:
2*η = λ⁶[ν₁ + ν₂]=
i*(λ²)*η = λ⁶[ν₁ + ν₂]
(e) Dividing both sides by λ² gives us:
i*η = λ⁴[ν₁ + ν₂]
(f) But this then implies that λ divides η which goes against assumption. [Note, it doesn't divide i because it is not a unit. It therefore divides η by Euclid's Lemma for Gaussian Integers]

QED

Gaussian Integers: properties of 1 - i

Today's blog continues a proof that was first presented in a previous blog. If you are new to unique factorization, start here. If you are new to Gaussian Integers, start here. To begin this proof, start here.

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

Today's blog focuses on properties of the Gaussian prime λ which we will need to prove Fermat's Last Theorem for n = 4.

Definion 1: λ is a Gaussian Integer that is equal to 1 - i.

Lemma 1: λ is a prime.

Since Norm(λ) = (1 - i)(1 + i) = 1 + 1 = 2. [See here for further details on why this proves that λ is a prime.]

QED

Lemma 2: λ divides 2

i * λ² = (i)(1 - i)² = (i)(1 -2i - 1) = 2.

QED

Corollary 2.1: λ⁶ divides 8

From Lemma 2 above,
(i*λ²)³ =(-i)*λ⁶= 8

QED

Lemma 3: λ*i = (1 + i)

λ * i = i(1 - i) = i + 1

QED

Definition 2: ≡ is relation such that α ≡ β (mod γ) means that α - β is divisible by γ.

We describe this relationship by saying that α modulus γ is equal to β

This definition is true for Gaussian Integers or standard integers. For example, we know that 6 ≡ 2 (mod 4) since 4 divides 6 - 2. We then can say that 6 modulus 4 is equal to 2.

Lemma 4: Let α be any Gaussian Integer. α modulus 2 equals 0, 1, i, or λ

(1) For any α, there exists a,b such that α = a + bi. [Definition of Gaussian Integer]

Case I: a is even, b is even

In this case, α ≡ 0 (mod 2) since a is even, b is even implies there exists A, B such that a = 2A, b = 2B and α = 2A + 2Bi = 2(A+Bi)

Case II: a is odd, b is even

In this case, α ≡ 1 (mod 2) since there exists A, B such that a = 2A+1, b = 2B and α - 1 = 2A + 1 - 1 + 2Bi = 2(A + Bi)

Case III: a is even, b is odd

In this case, α ≡ i (mod 2) since there exists A, B such that a = 2A, b = 2B + 1 and α - i = 2A + (2B+1)i - i = 2A + 2Bi = 2(A+Bi)

Case IV: a is odd, b is odd

In this case, α ≡ λ (mod 2) since there exists A,B such that a = 2A + 1, b = 2B + 1, and α - λ = 2A + 1 + (2B + 1)i - (1 - i) = 2A + 1 + 2Bi + i + i - 1 = 2A + 2B + 2i = 2(A + B + i)

QED

Lemma 5: if α is a Gaussian Integer that is not divisible by λ, then α⁴ ≡ 1 (mod λ⁶).

(1) Since α is not divisible by λ, it cannot be divisible by 2. [See Lemma 2 above]

(2) We also know that modulo 2, it cannot be λ. If α modulo 2 is λ then it would imply that α is divisible by λ which it is not.

(3) So, we are left with α ≡ 1 or i (mod 2) from Lemma 4.

(4) α⁴ ≡ 1 (mod 8) since:

Case I: α ≡ 1 (mod 2)

(a) (α + 1) ≡ (α - 1) ≡ 0 (mod 2)

(b) (α² - 1) ≡ (α + 1)(α - 1) ≡ 0 (mod 4)

(c) (α² + 1) ≡ 0 (mod 2)

(d) (α⁴ - 1) ≡ (α² + 1)(α² - 1) ≡ 0 (mod 8)

Case II: α ≡ i (mod 2)

(a) (α + i) ≡ (α - i) ≡ 0 (mod 2)

(b) (α² + 1) ≡ (α + i)(α - i) ≡ 0 (mod 4)

(c) (α² - 1) ≡ 0 (mod 2)

(d) (α⁴ - 1) ≡ (α² + 1)(α² - 1) ≡ 0 (mod 8)

(5) And this proves that α⁴ ≡ 1 (mod λ⁶) using Corollary 2.1 above.

QED