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Perfect Square (Posted on 2006-05-07) Difficulty: 4 of 5
Determine the set of integers for which n^2 + 19n + 92 is a square.

No Solution Yet Submitted by Ravi Raja    
Rating: 2.0000 (2 votes)

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Solution solution | Comment 1 of 2

The solution to this puzzle is as follows.
By the problem,
n^2 + 19n + 92 = p^2, where p is an integer whether positive or negative.
Or, (2n+19)^2 =(2*p)^2 -7
Or, (2p+2n+19)(2p-2n-19) =7*1=1*7 = -7*-1=-1*-7; giving:
(2p+2n+19, 2p-2n-19)=(7,1),(1,7),(-7,-1),(-1,-7)
Solving, we obtain: (p, n) = (2, -8);(2,-11);(-2, -11);(-2,-8); so that  the only integers   satisfying the equation under reference are given by   n = -8  and  n = -11.
(It is never mentioned in the problem that n must be positive)

Edited on May 7, 2006, 9:46 am
  Posted by K Sengupta on 2006-05-07 09:45:45

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