Which of the following integer values of n produce a perfect integer square for n^2+9n+14?

×
1

✓
2

×
5

×
36
Which of the following integer values of n produce a perfect integer square for n^2+9n+14?
1
2
5
36
Let m\in Z such that
n^2+9n+14=m^2,
A perfect square number can be expressed by a smaller perfect number plus an integer, which can be further expressed as the product of two inters using the difference of two squares.
4n^2+36n+4\times 14=4m^2
4n^2+36n+81+4\times 14=4m^2+81
(2n+9)^24m^2=8156 = 25
(2n+92m)(2n+92m) = 25
Let a = 2n+92m, b = 2n+92m
Addition of above equations gives
n = \dfrac{a+b18}{4}
and
a\cdotp b =1\times 25=5\times 5 = 25
When a = 1, b=25,
n =\dfrac{2618}{4} = 2
n^2+9n+14 = 2^2+9\times 2+14=36
m = 6
When
So B is the choice