A proof journey · Diophantine equations
Right triangles lead to infinite descent.
First describe every integer right triangle. Then turn the description around: a hypothetical solution of a quartic or cubic equation would force a smaller one, forever.
What the method must deliver
Infinite descent is the well-ordering principle used backwards. Assume a solution exists, select one minimizing a specified positive integer, and construct another solution with a strictly smaller value of that integer. A finite search cannot replace this last step.
Here “biquadratic” means the fourth-power equation x⁴ + y⁴ = z²; “cubic” means the Fermat equation x³ + y³ = z³. These are particular Diophantine equations, not claims that all equations of degree four or three have no solutions.
Prerequisites: gcd, Bézout, prime factorization, and congruences. The cubic proof needs a new ring; its norm, units, and factorization property are developed below. Read the triple and quartic sections first.
1. Every primitive Pythagorean triple
A positive triple (a,b,c) satisfies a² + b² = c². It is primitive if gcd(a,b,c) = 1; then every pair is coprime. Two odd legs would give c² ≡ 2 modulo 4, so exactly one leg is even. Label that leg b.
m > n > 0, gcd(m,n) = 1, m and n of opposite parity.
Classification theorem. These formulas give every primitive positive triple, uniquely when the even leg is second. Every positive triple is a positive integer multiple of a primitive one.
Why the parameters exist
Both a and c are odd. Put r = (c+a)/2 and s = (c−a)/2. These are positive integers, and rs = (b/2)². A common divisor of r and s divides their sum c and difference a; hence gcd(r,s) = 1.
By unique factorization, a product of coprime positive integers is a square only when each factor is a square. Write r = m² and s = n². Adding and subtracting gives c and a above; positivity gives b = 2mn. Coprimality of r,s gives gcd(m,n) = 1, and c odd forces opposite parity. The values m = √r and n = √s show uniqueness.
Conversely, the formulas satisfy a²+b²=c² by expansion. An odd prime dividing both legs would divide both m and n, impossible. The odd leg excludes a common factor 2. Thus the stated conditions make the triple primitive.
Build a triple. Test the hypotheses.
Parameters coprime: yes. Opposite parity: yes. Both conditions are needed for a primitive triple. The identity a² + b² = c² holds for every permitted pair.
Try (m,n) = (3,1), then (4,2). The first fails parity; the second fails coprimality. Dividing the resulting triple by its common divisor recovers a primitive one.
A worked example and a warning
For (m,n) = (3,2), the triple is (5,12,13). Conversely, (c+a)/2 = 9 and (c−a)/2 = 4 recover the parameters. For (3,1), the formulas give (8,6,10), which is twice (4,3,5). The identity still holds, but the parity hypothesis for primitiveness fails.
2. The biquadratic equation: descend on z
Theorem. There are no positive integers x,y,z with x⁴ + y⁴ = z². This is stronger than the exponent-four Fermat theorem: a solution x⁴+y⁴=w⁴ would give z=w².
Assume a smallest solution
- Normalize. Choose a solution with z least. If d = gcd(x,y) > 1, then d² divides z by prime factorization, and dividing by d⁴ gives a smaller solution. Thus gcd(x,y)=1. Both x,y odd would give z² ≡ 2 modulo 16. Relabel so x is odd and y is even.
- Use the triple theorem once. The primitive triple (x²,y²,z) gives x²=u²−v², y²=2uv, z=u²+v², with u>v>0, gcd(u,v)=1, and opposite parity. Here u is odd and v even: the other choice would make x² ≡ −1 modulo 4.
- Use it a second time. Now (x,v,u) is itself a primitive right triangle, since x²+v²=u² and any common prime would also divide u and v. Therefore x=s²−t², v=2st, u=s²+t², with s>t>0, gcd(s,t)=1, and opposite parity.
- Separate the squares. Substitution gives (y/2)²=st(s²+t²). The three positive factors s, t, s²+t² are pairwise coprime. For example gcd(s,s²+t²)=gcd(s,t²)=1. Thus each is a square: s=A², t=B², s²+t²=C².
- Exhibit the smaller solution. We have A⁴+B⁴=C², with A,B,C positive. Moreover C²=u and z=u²+v², so C<z. This contradicts the choice of z.
The descent does not merely produce a smaller parameter inside an unrelated triangle. It produces a new solution of the same fourth-power equation, and compares its right-hand variable directly with the original one.
3. A ring suited to cubes
The real factorization x³+y³=(x+y)(x²−xy+y²) hints at a quadratic extension. Let ω be a nonreal cube root of 1, so ω²+ω+1=0. The Eisenstein integers are R=ℤ[ω]={a+bω : a,b∈ℤ}. Define λ=1−ω.
R×={±1, ±ω, ±ω²}, N(λ)=3, 3=−ω²λ².
Why unique factorization is available
The norm is the ordinary squared complex absolute value, so it is positive on nonzero elements and multiplicative. Solving N(a+bω)=1 gives exactly the six units listed above. Given α/β in the complex plane, round its two coordinates in the basis 1,ω to integers. The remaining coordinates u,v lie in [−1/2,1/2], so u²−uv+v²≤3/4<1. Multiplying back gives α=qβ+r with N(r)<N(β).
Thus R is Euclidean. The Euclidean algorithm gives gcds and Bézout; irreducibles are prime, and the usual proof gives unique factorization up to units. Since λ has prime norm 3, it is prime. Reduction modulo λ identifies ω with 1 and gives R/(λ)≅𝔽₃. Write vλ(α) for the exponent of λ in a nonzero α.
Two traps to avoid. The factors α+β, α+ωβ, α+ω²β will share λ. Also, the cubes of the six units are only ±1: an arbitrary unit cannot simply be absorbed into a cube.
4. Cubic descent: decrease a prime exponent
Theorem. There are no nonzero integers x,y,z satisfying x³+y³=z³. Signed variables are allowed; zero gives trivial solutions and is excluded.
We prove a stronger statement inside R. There are no pairwise coprime nonzero α,β,γ, each prime to λ, satisfying the following equation for a unit ε and a positive integer m:
If any such equation exists, choose one with m least. The descent will reduce m to m−1.
Step 1. A small congruence rules out m=1
Since λ² is associated to 3, R/(λ²) has nine classes (the two integer coefficients taken modulo 3); six are prime to λ. The six units give distinct classes there, since the difference of two distinct units has norm 1, 3, or 4, never a positive multiple of N(λ²)=9. Hence any λ-free α has the form e+λ²t for a unit e. Cubing and using vλ(3)=2 shows α³≡e³=±1 modulo λ⁴.
The same holds for β. Their signs must cancel, because λ divides α³+β³ whereas λ does not divide 2. Thus λ⁴ divides their sum. The right side has valuation exactly 3m, giving 3m≥4 and therefore m≥2.
Step 2. Factor, keeping the shared prime
Put Fi=α+ωiβ for i=0,1,2. Their product is α³+β³. Each is divisible by λ. Any common prime other than λ would divide both α and β by subtracting two factors; this contradicts coprimality. Also vλ(Fi−Fj)=1, so at most one factor has valuation at least 2. At least one does, since the three valuations sum to 3m≥6.
Replace β by ωβ or ω²β if necessary; its cube stays the same and the factors rotate. We may take F₀ to be the exceptional factor. Unique factorization now gives units e₀,e₁,e₂ and pairwise coprime λ-free nonzero g₀,g₁,g₂ with
F₁=e₁ λ g₁³, F₂=e₂ λ g₂³.
Every other prime exponent in a factor is divisible by 3, because such a prime occurs in only that factor and the product is a unit times λ3m times a cube.
Step 3. Recombine and control the unit
The identity F₀+ωF₁+ω²F₂=0, divided by ωe₁λ, becomes
η=ωe₂/e₁, θ=−e₀/(ωe₁).
Both η and θ are units. Reduce modulo λ³. Since m≥2 the right side vanishes; since the g’s are λ-free, their cubes are ±1 modulo λ³ by Step 1. Thus η≡±1 modulo λ³. Among the six units, only +1 can be congruent to +1, and only −1 to −1, modulo λ³: differences of distinct units have norm at most 4, whereas N(λ³)=27. Hence η is actually ±1.
Absorb that sign into g₂. We have a new equation of precisely the assumed kind, with the pairwise coprime λ-free entries g₁, ±g₂, g₀ and positive exponent m−1. This contradicts minimality. The stronger statement is proved.
Step 4. Return to ordinary integers
A primitive signed solution can be written x³+y³+w³=0. It is pairwise coprime: an ordinary prime dividing two variables would divide the third. Cubes modulo 9 are 0,1,−1. Three nonzero residues cannot add to 0 modulo 9; and two variables divisible by 3 would force all three to be. Thus exactly one variable is divisible by 3. Move it to the right, absorbing a sign, and write that variable 3kv with k≥1 and 3∤v.
The other two variables and v are pairwise coprime and λ-free. Their ordinary Bézout identities also establish coprimality in R. Since 3=−ω²λ², the equation becomes α³+β³=ε λ6kv³, the forbidden stronger equation with m=2k.
Compare the two descents. The quartic argument shrinks a positive integer appearing in a solution. The cubic argument shrinks the exponent of the ramified prime λ in an enlarged family of equations. Enlarging the family is what allows the descent to close.
5. Try the arguments yourself
1. Recover the parameters of (20,21,29), putting the even leg second.
Use (a,b,c)=(21,20,29). Then (c+a)/2=25 and (c−a)/2=4, so (m,n)=(5,2). Both parameters are coprime and of opposite parity.
2. Why must a primitive triple have exactly one even leg? Show that its even leg is divisible by 4.
Two even legs violate primitiveness; two odd legs give c²≡2 modulo 4. In b=2mn, exactly one parameter is even, so 4 divides b.
3. Find all primitive positive triples with hypotenuse at most 30.
Up to exchanging legs: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29). Enumerate coprime, opposite-parity m>n with m²+n²≤30; the triple theorem proves completeness.
4. In the quartic proof, justify each use of “coprime product is a square.”
For the triple theorem, gcd((c+a)/2,(c−a)/2)=1. In the descent, gcd(s,t)=1 implies gcd(s,s²+t²)=gcd(t,s²+t²)=1. Prime exponents in each individual factor are therefore even.
5. Explain why N(λ)=3 does not mean that 3 is prime in R.
λ has prime norm and is a prime element; the integer 3 is a unit times λ². The rational prime 3 ramifies in this larger ring.
6. Expand (a+bω)³ and identify the coefficient of ω.
Using ω²=−1−ω and ω³=1 gives (a+bω)³=a³−3ab²+b³+3ab(a−b)ω. The coefficient is always divisible by 3.
7. Identify the new solution and the decreasing integer in each descent.
Quartic: (A,B,C) solves A⁴+B⁴=C² and C<z. Cubic: (g₁,±g₂,g₀) solves the generalized unit equation with m replaced by m−1. Merely claiming that the factors are “smaller” would not prove either descent.
8. Why does an unsuccessful search up to one million prove neither theorem?
It excludes only a finite region. Descent excludes every solution by showing that existence would contradict the well-ordering of the positive integers.