Worked solution: Mihăilescu's proof of Catalan's conjecture via cyclotomic fields (2002)

Frequently asked questions

Why does it suffice to rule out odd prime exponents, rather than all exponents

u,v

Because if $x^u-y^v=1$ had a solution with composite $u$ (say $u=u'p$ ), then $(x^{u'})^p - y^v=1$ would give a solution with prime exponent $p$ too. So any counterexample to Catalan's conjecture, after this reduction and Lebesgue/Ko Chao's results, would already show up with $u=p$ and $v=q$ both odd primes.

Source

Preda Mihăilescu, 2002

Primary cyclotomic units and a proof of Catalan's conjecture

Condensed summary

Step 1 of 7: Catalan's 1844 conjecture and its reduction to two odd primes

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In plain words

Powers of whole numbers are usually spread far apart — $8=2^3$ and $9=3^2$ are consecutive, but that seems like a lucky accident. In 1844 the Belgian mathematician Eugène Catalan wrote to a journal conjecturing that $8$ and $9$ are the only pair of consecutive perfect powers, i.e. $x^u-y^v=1$ has no other solution with $x,y,u,v$ all greater than $1$ .

Before attacking the full conjecture, two classical results narrow the target: Victor-Amédée Lebesgue showed in 1850 that the exponent $2$ can never appear as $v$ (no equation $x^u-y^2=1$ has solutions for $u>1$ ), and Ko Chao showed in 1965 that $u=2$ forces the trivial solution. Together these reduce Catalan's conjecture to a single sharper statement about odd prime exponents.

x^u - y^v = 1,\quad x,y>0,\ u,v>1\ \implies\ 3^2-2^3=1

Detailed analysis

Catalan's 1844 note to Crelle's Journal conjectured that $x^u-y^v=1$ in integers $x,y,u,v>1$ has only the solution $3^2-2^3=1$ . The conjecture reduces in two classical steps. Victor-Amédée Lebesgue (1850) proved $x^u-y^2=1$ has no solutions for $u>1$ : factoring $y^2+1=(y+i)(y-i)$ in the Gaussian integers $\mathbb{Z}[i]$ , whose two factors are coprime because their difference divides $2i$ while $y$ must be even, forces each factor to be a $u$ -th power up to units, and comparing imaginary parts produces a descent contradiction. Ko Chao (1965) settled the mirror equation $x^2-y^v=1$ — the unique solution being $3^2-2^3=1$ — by congruence arguments that reduce it to cases already handled by Lebesgue and Nagell.

With both exponent-2 cases gone, the entire conjecture is equivalent to the single statement that $x^p-y^q=1$ has no solution in nonzero integers $x,y$ and odd primes $p,q$ (Bilu 2004, §1). This reduced equation carries an exact symmetry exploited throughout the proof: if $(x,y,p,q)$ is a solution, then so is $(-y,-x,q,p)$ , since $(-y)^q-(-x)^p=1$ expands, for odd $p,q$ , to $x^p-y^q=1$ .

The symmetry is not cosmetic — it is used constantly. Every divisibility, congruence, or exclusion proved for the ordered pair $(p,q)$ applies for free to $(q,p)$ : Cassels' relations of Step 1, the two Wieferich congruences of Step 3, and the exclusion $p\not\equiv1\pmod q$ of Step 4 all come in mirrored pairs, each proved once and invoked twice. Mihăilescu's argument (announced April 2002, published as 'Primary cyclotomic units and a proof of Catalan's conjecture', J. reine angew. Math. 572 (2004), and exposited by Bilu, Séminaire Bourbaki 2002–03, Astérisque 294) is directed entirely at this reduced symmetric equation.

Terms in this step

Catalan's equation: The equation $x^p-y^q=1$ in non-zero integers $x,y$ and odd primes $p,q$ — the sharpened, symmetric form to which Catalan's original conjecture reduces once the exponent- $2$ cases are handled separately.