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416 Chapter 8 THE UBIQUITY OF PRIME NUMBERS<br />

Observe that in seeking Diophantine solutions here we are simply addressing<br />

the problem of whether there exist higher instances of consecutive powers.<br />

An accessible treatment of the history of the Catalan problem to the date of<br />

its publication is [Ribenboim 1994], while more recent surveys are [Mignotte<br />

2001] and [Metsänkylä 2004]. Using the theory of linear forms of logarithms<br />

of algebraic numbers, R. Tijdeman showed in 1976 that the Catalan equation<br />

has at most finitely many solutions; in fact,<br />

y q 10 7<br />

and max{p, q} < 7.78 × 10 16 . Further, explicit easily checkable criteria on<br />

allowable exponent pairs were known, for example the double Wieferich<br />

condition of Mihăilescu: if p, q are Catalan exponents other than the pair<br />

2, 3, then<br />

p q−1 ≡ 1(modq 2 ) and q p−1 ≡ 1(modp 2 ).<br />

It was hoped that such advances together with sufficiently robust calculations<br />

might finish off the Catalan problem. In fact, the problem was indeed finished<br />

off, but using much more cleverness than computation.<br />

In [Mihăilescu 2004] a complete proof of the Catalan problem is presented,<br />

and yes, 8 and 9 are the only pair of nontrivial consecutive powers. It is<br />

interesting that we still don’t know whether there are infinitely many pairs of<br />

consecutive powers that differ by 2, or any other fixed number larger than 1,<br />

though it is conjectured that there are not. In this regard, see Exercise 8.20.<br />

Related both to Fermat’s last theorem and the Catalan problem is the<br />

Diophantine equation<br />

x p + y q = z r , (8.2)<br />

where x, y, z are positive coprime integers and exponents p, q, r are positive<br />

integers with 1/p+1/q+1/r ≤ 1. The Fermat–Catalan conjecture asserts that<br />

there are at most finitely many such powers x p ,y q ,z r in (8.2). The following<br />

are the only known examples:<br />

1 p +2 3 =3 2<br />

2 5 +7 2 =3 4 ,<br />

13 2 +7 3 =2 9 ,<br />

2 7 +17 3 =71 2 ,<br />

3 5 +11 4 = 122 2 ,<br />

33 8 + 1549034 2 = 15613 3 ,<br />

1414 3 + 2213459 2 =65 7 ,<br />

(p ≥ 7),

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