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 Looking for n (Posted on 2004-02-24)
Let n be the smallest positive integer such that n(n+1)(n+2)(n+3) can be expressed as either a perfect square or a perfect cube (not necessarily both).

Find n, or prove that this is not possible.

 Submitted by Aaron Rating: 4.2857 (7 votes) Solution: (Hide) It is not possible in either case. Proof: (n+1)(n+2) = n² + 3n + 2 and n(n+3) = n² + 3n. So their product is (n2 + 3n + 1)² - 1. Hence n(n + 1)(n + 2)(n + 3) is 1 less than a square, so it cannot be a square. One of n+1, n+2 must be odd. Suppose it is n+1. Then n+1 has no factor in common with n(n + 2)(n + 3), so n(n + 2)(n + 3) = n³ + 5n² + 6n must be a cube. But (n + 1)³ = n³ + 3n² + 3n + 1 < n³ + 5n² + 6n < n³ + 6n² + 12n + 8 = (n + 2)³, so n³ + 5n² + 6n cannot be a cube. Similarly, suppose n + 2 is odd. Then it has no factor in common with n(n + 1)(n + 3) = n³ + 4n² + 3n, so n³ + 4n² + 3n must be a cube. But for n >= 2, (n + 1)³ < n³ + 4n² + 3n < (n + 2)³, so n³ + 4n³ + 3n cannot be a cube for n >= 2. The case n = 1 is checked by inspection: 24 is not a cube.

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 Subject Author Date Puzzle Solution K Sengupta 2022-09-04 23:41:04 Solution Math Man 2011-08-22 10:57:10 1 less than a perfect square anil 2004-04-06 19:28:35 looking in the mirror, yo yo Penny 2004-03-23 11:17:15 Not a cube Nick Hobson 2004-02-29 11:00:33 part of solution red_sox_fan_032003 2004-02-27 12:35:23 not a square red_sox_fan_032003 2004-02-25 17:42:49 think in terms of factors tan 2004-02-25 02:36:55 re: I think i have one, oh no i haven't Juggler 2004-02-24 19:30:00 I think i have one Juggler 2004-02-24 19:22:44 re: Not a square SilverKnight 2004-02-24 15:46:53 Solution (?) Brian Smith 2004-02-24 15:36:24 Not a square Federico Kereki 2004-02-24 15:20:46
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