Determine the set of integers for which n^2 + 19n + 92 is a square.
n^2+19n+92=m^2 has solutions n=9 1/2 +/ (1/2)sqrt(4m^27). Thus
4m^27=k^2 where k is an integer or else n would be irrational. Thus
the difference (2m)^2k^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 20060507 16:41:37 