August 30th, 2012, 07:09 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, 09:35 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, 10:15 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, 02:53 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, 02:39 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, 03:49 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, 03:56 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: squares
Let be the base of the number system. Then the integer 14641 is represented by: 

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