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 * λ4r * 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'*λr where r ≥ 1.
(8) Let ε be some unit such as 1, -1, i, or -i. We can now state that:
ε * α'4 * λ4*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 * λ4*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 * λ4*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 λ4*r. [Note, ε is a unit and is only divisible by other units]
(22) And this shows that r ≥ 2.
QED
Saturday, June 25, 2005
Wednesday, June 22, 2005
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
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
Sunday, June 19, 2005
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
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
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