Consecutive integers which are perfect powers

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Mihăilescu's theorem

In 1844, Eugène Charles Catalan proposed the conjecture (proved in 2002 by Preda Mihăilescu, hence Mihăilescu's theorem)

Conjecture (Catalan's conjecture, 1844). (Catalan)

8 and 9 are the only consecutive integers which are perfect powers,

$3^{2}-2^{3}=1.\,$

Interestingly, 2 and 3 are the only consecutive primes.

Consecutive perfect powers with difference equal to k

A001597 Perfect powers: $\scriptstyle m^{k}\,$ where $\scriptstyle m\,$ is an integer and $\scriptstyle k\,\geq \,2.\,$

{1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 81, 100, 121, 125, 128, 144, 169, 196, 216, 225, 243, 256, 289, 324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1000, 1024, 1089, ...}

Consecutive perfect powers with difference equal to

1: (8, 9).

2: (25, 27), ?

3: (1, 4), (125, 128), ?

4: (4, 8), (32, 36), (121, 125), ?

5: (27, 32), ?

It seems that all those lists are finite. (CONJECTURE? or PROOF?)

See also

• Consecutive integers which are not squarefree