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Mihăilescu’s Theorem (Mihăilescu’s theorem)

Recommended reading: 【Number Theory】 Number Theory Table of Contents

1. Overview

2. Step 1. Kummer’s Attempt

3. Step 2. Double Wieferich Condition

4. Step 3. Pigeonhole Principle

5. Step 4. Thaine’s Theorem

6. Step 5. Proof by Contradiction

1. Overview

⑴ Catalan’s Conjecture

For integers $x,y,a,b>1$, the only solution satisfying $x^a-y^b=1$ is $3^2-2^3=9-8=1$.

⑵ It was conjectured by Eugène Catalan in 1844 and proved by the Romanian mathematician Preda Mihăilescu in 2002.

2. Step 1. Kummer’s Attempt

⑴ 1-1. Simplifying the Problem

① Assume that a hypothetical solution $x^m-y^n=1$ exists. Choose primes $p\mid m$ and $q\mid n$, and set $X=x^{m/p}$ and $Y=y^{n/q}$. Then we obtain a solution of the form $X^p-Y^q=1$.

② If one of the exponents is $2$, classical results already show that only $3^2-2^3=1$ is possible.

③ Therefore, if Catalan’s conjecture had a new solution, we could assume that $p$ and $q$ in $x^p-y^q=1$ are distinct odd primes.

⑵ 1-2. Kummer’s Attempt: Although it failed to prove Fermat’s Last Theorem, it provided a clue for Mihăilescu’s proof.

① $x^p-y^q=1\iff x^p-1=y^q$

○ Over the integers, it can only be factored as $x^p-1=(x-1)(x^{p-1}+\cdots+1)$.

○ However, in the cyclotomic field $\mathbb{Q}(\zeta)$ containing a primitive $p$-th root of unity $\zeta=\zeta_p$, it factors as

② Just as a product of pairwise coprime integers can be a $q$-th power only if each factor is a $q$-th power, after suitable normalization, the ideal of each factor must be a $q$-th power.

○ $\displaystyle \beta_i=\frac{x-\zeta^i}{1-\zeta^i}$

○ Since the ideals $(\beta_i)$ are pairwise coprime, $(\beta_i)=\mathfrak{a}_i^q$.

○ Stickelberger’s theorem: Since the ideal is a $q$-th power, an element generating the ideal is a unit multiple of a $q$-th power.

○ Roughly, this can be understood as $\beta_i=\varepsilon\alpha_i^q$, where $\varepsilon$ is a unit, but the precise relation is $\beta^\theta=\varepsilon\alpha^q$.

○ $\displaystyle \theta=\sum_i a_i\sigma_i$: This may be understood as analogous to an angle on the unit circle; see Step 3.

3. Step 2. Double Wieferich Condition

⑴ Overview

① $p$-th power of an ideal $\rightarrow$ double Wieferich condition

② Fermat’s little theorem, a special case of Euler’s totient theorem: $p^{q-1}\equiv1\pmod q$

③ The double Wieferich condition strengthens the modulus to $q^2$: $p^{q-1}\equiv1\pmod{q^2}$

⑵ 2-1. Lifting Phenomenon: In a cyclotomic field, if $n^q-m^q\equiv0\pmod{\mathfrak{Q}}$, then $n^q-m^q\equiv0\pmod{\mathfrak{Q}^2}$.

① Overview: For a given counterexample to Catalan’s conjecture, assuming $x\equiv0\pmod q$ gives $x^2\equiv0\pmod{q^2}$.

② For any sequence $a_i$ defining $\theta$,

because all terms of degree at least $2$ in $x$ vanish modulo $q^2$.

③ Similarly,

④ Subtracting ② from ③ gives

⑤ The difference of two $q$-th powers appears in ④ because the many available degrees of freedom allow us to understand, intuitively,

⑥ Stickelberger’s theorem: The coefficients can be chosen so that $a_i-a_{p-i}$ is not divisible by $q$. Therefore, ④ implies $x\equiv0\pmod{q^2}$.

⑶ 2-2. Derivation of the Wieferich Congruence

① Cassels’ relation: There exists an integer $u$ such that $x-1=p^{q-1}u^q$.

② By Fermat’s little theorem,

③ By the lifting phenomenon, ② gives $u^q\equiv-1\pmod{q^2}$.

④ Returning to ①,

⑤ Since Cassels’ relations are symmetric in $p$ and $q$, we similarly obtain $q^{p-1}\equiv1\pmod{p^2}$.

4. Step 3. Pigeonhole Principle

⑴ Overview: Impose bounds on $p$ and $q$.

⑵ 3-1. Lemma: For $\alpha$ arising from a small nonzero $\theta$,$\left|\arg(\sigma(\alpha))\right|>\pi/q$.

① $\displaystyle \theta=\sum_i a_i\sigma_i$

② Each $\theta$ gives a relation of the form $(x-\zeta)^\theta=\alpha_\theta^q$.

③ Intuitive interpretation: Since primes exhibit periodicity, distinct residues modulo $q$ must correspond to distinguishable angles associated with modulo $q$.

⑶ 3-2. Assuming $q>4p^2$ leads to a contradiction (ref): The constant $4$ is not a profound natural constant but rather quantitative slack in the proof.

① Overview: As $q$ increases, the admissible range for $\theta$ also increases, and the number of $\theta$ within that range in the high-dimensional lattice grows much faster than $q$.

② Basis of the Stickelberger ideal: $e_1,\ldots,e_r$, where $\displaystyle r=\frac{p-1}{2}$

③ $\theta=\lambda_1e_1+\cdots+\lambda_re_r$, where each $\lambda_i$ is a nonnegative integer.

④ The number of integer lattice points satisfying $\lambda_1+\cdots+\lambda_r\le s$ is

⑤ When $K=4$, we obtain $s\ge6$, so

⑥ Therefore, within a sufficiently large range of $p$, we have $N>q$.

⑦ Because primes exhibit periodicity, all lattice points would need to have distinct residues modulo $q$ to remain distinguishable from the perspective of $q$. By the pigeonhole principle, at least two have the same residue modulo $q$. Therefore, no counterexample to Catalan’s conjecture can satisfy $q>4p^2$.

5. Step 4. Thaine’s Theorem

⑴ 4-1. If a hypothetical solution to Catalan’s equation exists, an abnormally large number of $q$-primary cyclotomic units behaving like $q$-th powers arise.

⑵ 4-2. The structure of the unit group and class group of the cyclotomic field, particularly Thaine’s theorem, shows that so many such units cannot exist.

⑶ Conclusion: $p\equiv1\pmod q$ or $q\equiv1\pmod p$

6. Step 5. Proof by Contradiction

⑴ Combining Steps 1–4, the exponents of a hypothetical solution must satisfy all four of the following conditions.

① $p,q\ge11$

② $p^{q-1}\equiv1\pmod{q^2}$ and $q^{p-1}\equiv1\pmod{p^2}$

③ $p<4q^2$ and $q<4p^2$

④ $p\equiv1\pmod q$ or $q\equiv1\pmod p$

⑵ Deriving a contradiction

① Without loss of generality, assume $p\equiv1\pmod q$ and write $p=kq+1$.

② Since

we obtain $k=Kq\equiv0\pmod q$.

③ Since $p=Kq^2+1<4q^2$, we must have $K\in{1,2,3}$.

④ If $K=1$ or $K=3$, then $p$ is even, contradicting the fact that $p$ is an odd prime.

⑤ If $K=2$, then $q^2\equiv1\pmod3$ gives

Thus, $p$ is divisible by $3$, contradicting $p\ge11$.

⑥ Therefore, Catalan’s equation has no solution other than $3^2-2^3=1$.

Entered: 2026.08.24 22:17