If $ x^4 + ax^3 + bx^2 + cx + 1 $ is the square of a polynomial, then it must be quadratic. We can assume that the quadratic is monic. Then to get a term of $ ax^3 $ when we square it, the coefficient of $ x $ in the quadratic must be $ \frac{a}{2} $. Hence, \[x^4 + ax^3 + bx^2 + cx + 1 = \left( x^2 + \frac{a}{2} \cdot x + t \right)^2.\]Expanding, we get \[x^4 + ax^3 + bx^2 + cx + 1 = x^4 + ax^3 + \left( \frac{a^2}{4} + 2t \right) x^2 + atx + t^2.\]Matching coefficients, we get \begin{align*} \frac{a^2}{4} + 2t &= b, \\ at &= c, \\ t^2 &= 1.\end{align*}Similarly, if $ x^4 + 2ax^3 + 2bx^2 + 2cx + 1 $ is the square of a polynomial, then we can assume the polynomial is of the form $ x^2 + ax + u $. Hence, \[x^4 + 2ax^3 + 2bx^2 + 2cx + 1 = (x^2 + ax + u)^2.\]Expanding, we get \[x^4 + 2ax^3 + 2bx^2 + 2cx + 1 = x^4 + 2ax^3 + (a^2 + 2u) x^2 + 2aux + u^2.\]Matching coefficients, we get \begin{align*} a^2 + 2u &= 2b, \\ 2au &= 2c, \\ u^2 &= 1.\end{align*}From the equations $ at = c $ and $ 2au = 2c, $ $ t = \frac{c}{a} = u $. Thus, we can write \begin{align*} \frac{a^2}{4} + 2t &= b, \\ a^2 + 2t &= 2b, \\ at &= c, \\ t^2 &= 1.\end{align*}Since $ t^2 = 1, $ either $ t = 1 $ or $ t = -1 $. If $ t = 1, $ then $ \frac{a^2}{4} + 2 = b $ and $ a^2 + 2 = 2b $. Substituting for $ b, $ we get \[a^2 + 2 = \frac{a^2}{2} + 4.\]Then $ a^2 = 4, $ so $ a = 2 $. Then $ b = 3 $ and $ c = 2 $. If $ t = -1, $ then $ \frac{a^2}{4} - 2 = b $ and $ a^2 - 2 = 2b $. Substituting for $ b, $ we get \[a^2 - 2 = \frac{a^2}{2} - 4.\]Then $ a^2 = -4, $ which has no real solutions. Therefore, $ a = 2, $ $ b = 3, $ and $ c = 2, $ so $ a + b + c = \boxed{7} $.