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 conjecture, as stated in his 1844 note in Crelle's journal, asserts that the equation $x^u-y^v=1$ has exactly one solution in integers $x,y>0$ and $u,v>1$ , namely $3^2-2^3=1$ (Bilu 2004, §1, quoting Catalan 1844). For 158 years this remained open, despite partial results (Lebesgue 1850 for $v=2$ ; Ko Chao 1965 for $u=2$ ).

Combining Lebesgue's and Ko Chao's theorems reduces the whole conjecture to: the equation $x^p-y^q=1$ has no solution in non-zero integers $x,y$ when $p,q$ are odd primes (Bilu 2004, §1, Conjecture 1.2). Because the equation is now symmetric in a useful sense — if $(x,y,p,q)$ solves it, so does $(-y,-x,q,p)$ — this reduced statement is the actual target of Mihăilescu's proof, and is what mathematicians now call Catalan's equation.

The remaining steps trace Mihăilescu's 2002/2004 argument (published in Crelle's journal, and surveyed by Yuri Bilu at the Séminaire Bourbaki, Astérisque 294, 2004) for why $x^p-y^q=1$ is impossible for odd primes $p,q$ .

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.