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2005 a square? Baseless! (Posted on 2005-05-30) Difficulty: 3 of 5
2005 base 10 is not a square. Neither is 2005 base 7 a square (equal 2*7^3+5=691). Is there any base b such that 2005 base b is a square?

See The Solution Submitted by McWorter    
Rating: 3.6000 (5 votes)

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Some Thoughts A Modulus Based Solution !?! | Comment 13 of 18 |
A quip was made by McWorter about other mathematical bases.
<p>
I wonder about the philosophical legitimacy of my solution, but within the limited parameters which I have defined for myself, my solution has validity.
<p>
If I take the number 4 (base 10) into a modular arithmetical/algebraic environment, eg mod(5), 4 still remains as 4.  It is still the square of 2.
<p>
Now I think I tread on eggshells:<br>
    I am considering 2005 base 10 under the modulus of 2001.  2001 => 0, 2002 => 1, 2003 => 2, 2004 => 3 and 2005 => 4!
<br> Yes, 2 * 2 = 4, and I have a square.
<p>
A generality of my approach to this is:   2005 (mod(2001-x^2)) = x * x.
<p> Now before any would like to drag me violently kicking to La Place de L'Etoile to exact penance as a heretic I will offer these considerations:<br>
1.  Back to base 10 mod(5).  2 * 2 = 4,  but 4 * 4 = 1.  [Now here might be another base to work from!]<br>
2.  In reality, the highest value that can exist for mod(x) is x-1.
<p> Good piece of trivial exploration?  But I guess that I've destroyed my own case!  Bonjour Madame Guillotine!<p>
<p> Hey! Sorry, this is probably a Forum matter, but if I get away with playing with squares in a modulus environment, then why not .... primes ... Fermat numbers ... Fibonacci series ...
<br> Some of these instances obvious come up in practical program situations, but philosophically, is this a legitimate tack? Yes, I will take it there.  Comments to that end, please go to Forums/General.
  Posted by brianjn on 2005-06-01 07:30:34
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