Find the value of an integer a such that a^2 +6a +1 is a perfect square.

Question

I was able to solve this but it required me using hit and trial at one step. I was wondering if i could find a more solid method to solve it.

p.s. this is the first time im asking a question here so sorry if i couldn't construct the question properly.

this is my solution -

a² + 6a + 1 = k²

(a+3)² = k² + 8

k² + 8 = m² [m=a+3]

(m-k)(m+k) = 8 ... here by inspection m= 3 and k =1

hence, a = 0, -6.

Answer 1

f(a) = a^2 + 6a+ 1

f(0) = 1 which is a perfect square and that solves the problem.

So when a=0, the function is a perfect square, namely 1.

Let's you want another , like 9

f(x) = 9

a^2 + 6a + 1 = 9

a^2 + 6a = 8

a^2 + 6a + 9 = 8+9 = 17 <--- completes the square

(a+3)^2 = 17 <--- solves for a

(a+3) = +-sqrt(17)

a = -3 +- sqrt(17)

Now, how about 100?

f(x)=100

a^2 + 6a + 1 = 100

a^2 + 6 a = 99

a^2 + 6a + 9 = 108

(a+3) ^2 = 108

a+3 = +-sqrt(108)

a+3 = +- sqrt(2 * 2 * 3 * 3 * 3)

= +- 2*3 * sqrt(3)

=+- 6*sqrt(3)

a = -3 +- 6*sqrt(3)