# Prime, followed by the cube of a prime, followed by the square of a prime. Other examples?

The numbers 7, 8, 9, apart from being part of a really lame math joke, also have a unique property. Consecutively, they are a prime number, followed by the cube of a prime, followed by the square of a prime. Firstly, does this occurence happen with any other triplet of consecutive numbers? More importantly, is there anyway to predict or determine when and where these phenomena will occur, or do we just discover them as we go?

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Hint : x^2 - 1 = (x-1)(x+1) – mick Oct 12 '12 at 14:06

Catalan's conjecture implies that a square and a cube cannot be consecutive unless they are (0,1) or (8,9). So this is the only triplet satisfying this property.

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There are also more elementary proofs known that a square and a cube are not consecutive (aside from those examples). This was apparently already known to Euler using descent. – Erick Wong Oct 12 '12 at 1:08
I just have to point out that this is now Mihăilescu's theorem :) which of course your link points out but is still worth having here for the lazy clicker. – kcrisman Oct 31 '12 at 1:13

Suppose the first prime is $x$ and $x+1=y^3$ where $y$ is also a prime. Then $y$ has be an odd number which implies $x$ is even. So $x$ cannot be a prime. Contradiction.

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Of course this exact pattern cannot recur as Patrick Li easily observed, but there are plenty of neat coincidences to be found in any group of small numbers. My favorite: of 243, 343, 443, the first is the fifth power of a prime, the second is the cube of a prime, and the third is a prime.

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If you'll settle for a prime, cube of a prime, square of a prime in arithmetic progression (instead of consecutive), you've got $$5,27=3^3,49=7^2\qquad \rm{(common\ difference\ 22)}$$ and $$157,\ 343=7^3,\ 529=23^2 \qquad \rm{(common\ difference\ 186)}$$ and, no doubt, many more where those came from. A bit more exotic is the arithmetic progression $81,\ 125,\ 169$ with common difference 44: the 4th power of the prime 3, the cube of the prime 5, the square of the prime 13. So $$3^4,5^3,13^2$$ is an arithmetic progression of powers of primes, and the exponents are also in arithmetic progression.

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