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Square An Integer, Get Quartic (Posted on 2007-05-13) Difficulty: 3 of 5
Analytically determine all possible integer pairs (p, s) such that:
4p4+ 4p3+ 4p2 + 4p +1 = s2.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts Partial solution | Comment 2 of 4 |
From the equation, we know that sē must be odd, since the term (4p4+ 4p3+ 4p2 + 4p) is even.  Therefore, s must also be odd.

We also know that for every odd s:       (s+1)(s-1) is evenly divisible by 4.

The obvious solutions are:

{(0,-1), (0,1), (2,-11), (2,11)}

However, as p increases, something ppeculiar happens.

(4p4+ 4p3+ 4p2 + 4p) = i + r

where i = an integer & r = the remainder.  The solution of course is to find a p where r = 0.  However, as p→∞, it appears that r→0.75.  If this holds true, then there are no more solutions for this equation.

  Posted by hoodat on 2007-05-14 13:25:19
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