#### 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