jmc

If $x^4+a{x}^3+b{x}^2+cx+1$ 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 $a{x}^3$ when we square it, the coefficient of $x$ in the quadratic must be $\frac{a}{2}.$ Hence, $$x^4+a{x}^3+b{x}^2+cx+1={\left(x^2+\frac{a}{2}\cdot x+t\right)}^2.$$Expanding, we get $$x^4+a{x}^3+b{x}^2+cx+1=x^4+a{x}^3+\left(\frac{a^2}{4}+2t\right)x^2+atx+t^2.$$Matching coefficients, we get $$\begin{array}{rl}\frac{a^2}{4}+2t& =b,\\ at& =c,\\ t^2& =1.\end{array}$$Similarly, if $x^4+2a{x}^3+2b{x}^2+2cx+1$ is the square of a polynomial, then we can assume the polynomial is of the form $x^2+ax+u.$ Hence, $$x^4+2a{x}^3+2b{x}^2+2cx+1=(x^2+ax+u{)}^2.$$Expanding, we get $$x^4+2a{x}^3+2b{x}^2+2cx+1=x^4+2a{x}^3+(a^2+2u)x^2+2aux+u^2.$$Matching coefficients, we get $$\begin{array}{rl}{a}^2+2u& =2b,\\ 2au& =2c,\\ u^2& =1.\end{array}$$From the equations $at=c$ and $2au=2c,$ $t=\frac{c}{a}=u.$ Thus, we can write $$\begin{array}{rl}\frac{a^2}{4}+2t& =b,\\ a^2+2t& =2b,\\ at& =c,\\ t^2& =1.\end{array}$$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 $a+b+c=\boxed{7}.$