Square Trouble

let $x^4+ax^3+bx^2+cx^2+1= (x^2+\alpha x+1)^2$ $x^4+ax^3+bx^2+cx^2+1=x^4+2\alpha x^3+(\alpha^2+2) x^2+2\alpha x+1$ we can say $\begin{cases} a=c=2\alpha\\ b=\alpha^2+2=\frac{a^2}{4}+2\end{cases}$ let $x^4+2ax^3+2bx^2+2cx^2+1=(x^2+\beta x+1)^2$ $x^4+2ax^3+2bx^2+2cx^2+1=x^4+2\beta x^3+(\beta^2+2) x^2+2\beta x+1$ we can say that $\begin{cases} a=\beta\\ 2b=\beta^2+2=a^2+2\end{cases}$ inserting b from the first expression, $2(\dfrac{a^2}{4}+2)=a^2+2\longrightarrow a=2$ and $a+b+c=a+a+\dfrac{a^2}{4}+2=2+2+\dfrac{4}{4}+2=\boxed{7}$