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 Square An Integer, Get Quartic (Posted on 2007-05-13)
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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 Squeeze Play (Solution) | Comment 3 of 4 |

If p satisfies the compound inequality (2p^2 + p + 1)^2 > 4p^4 + 4p^3 + 4p^2 + 4p + 1 > (2p^2 + p)^2 then it is obvious that the middle expression is not a square since it is strictly between two consecutive squares.

The only intervals where the compound inequality fails are [-1, -1/3] and [0, 2].  There are only four integers in this range: -1, 0, 1, and 2.

Direct calculation shows that p=1 is not a solution and p=-1 and p=0 are solutions with s = +/-1 and p=2 is a solution with s=+/-11.

The complete solution set is (-1, 1), (-1, -1), (0, 1), (0, -1), (2, 11), (2, -11).

 Posted by Brian Smith on 2007-05-14 23:02:03

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