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 August 30th, 2012, 08:09 AM #1 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 squares If any base greater than 6 is used, 14,641 is a perfect square, more precisely prove that 14641 is the fourth power of an integer in any base greater than 6. How do you prove this?
 August 30th, 2012, 10:35 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: squares In base 10, 14,641 is 1 * 10^4 + 4 * 10^3 + 6 * 10^2 + 4 * 10^1 + 1 * 10^0. What is it in base b? Can you factor that?
 August 30th, 2012, 11:15 AM #3 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: squares Here is another hint: This is the first few rows of Pascal's Triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 The number 121 is a perfect square in any base greater than 2, and the number 1331 is a perfect cube in any base greater than 3.
 August 30th, 2012, 03:53 PM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: squares Ooh... nice hint, but probably harder to get from that to the answer than from the answer to that!
 September 1st, 2012, 03:39 AM #5 Math Team   Joined: Apr 2012 Posts: 1,579 Thanks: 22 Re: squares I'd like to see the generalized Pascal thing proven. Sounds interesting.
 September 1st, 2012, 04:49 AM #6 Senior Member   Joined: Sep 2011 Posts: 140 Thanks: 0 Re: squares Sorry, I don't get this at all...... can someone give me a solution to this?... I'm really bad at this.
 September 1st, 2012, 04:56 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: squares Let $6 be the base of the number system. Then the integer 14641 is represented by: $1\cdot b^4+4\cdot b^3+6\cdot b^2+4\cdot b+1=\sum_{k=0}^4{4 \choose k}\cdot b^{4-k}\cdot1^k=(b+1)^4$

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# 1331 is a perfect cube in which od tge following bases

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