 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Perfect Square (Posted on 2006-05-07) 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) Comments: ( Back to comment list | You must be logged in to post comments.) Straightforward Method Comment 2 of 2 | n^2+19n+92=m^2 has solutions n=-9 1/2 +/- (1/2)sqrt(4m^2-7). Thus 4m^2-7=k^2 where k is an integer or else n would be irrational. Thus the difference (2m)^2-k^2 of two squares must be 7, but successive squares j^2 and (j+1)^2 differ by 2j+1 and thus the only two squares that differ by 7 are 16 and 9, giving us m=2 and k=3. Thus n can only be -9 1/2 +/- 1 1/2 = -8 or -11.
 Posted by Richard on 2006-05-07 16:41:37 Please log in:
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