Question
Find all possible solutions such that n^2+9n-3 is a perfect integer square.
Find all possible solutions such that n^2+9n-3 is a perfect integer square.
Let m be the integer
n^2+9n-3 = m^2
4n^2+36n-12 = 4m^2
4n^2+36n-12 +81= 4m^2+81
Using the formula for difference of two squares,
(2n+9)^2-4m^2 = 93
(2n+9-2m)(2n+9+2m) = 93
设 a = 2n+9-2m, b = 2n+9+2m,
Addition of two equations yields:
n =\dfrac{a+b-18}{4}
and
a\cdotp b =93
93 = 1\times 93 = 3\times 31 ,
that is a,b = 1,93 or 3,31
n = \dfrac{1+93-18}{4} =\dfrac{76}{4} = 19
m^2 = 19^-9\times 19-3 = 529
\therefore m =23
or
n = \dfrac{3+31-18}{4} = 4
m^2 = 4^2-9\times 4-3 = 49
\therefore m = 7