jmc

Let $f(x)=a{x}^4+b{x}^3+c{x}^2+bx+a$. Thus, the problem asserts that $x=2+i$ is a root of $f$.

Note the symmetry of the coefficients. In particular, we have $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ for all $x\ne 0$. Thus, if $x=r$ is any root of $f(x)$, then $x=\frac{1}{r}$ is also a root.

In particular, $x=\frac{1}{2+i}$ is a root. To write this root in standard form, we multiply the numerator and denominator by the conjugate of the denominator: $$\frac{1}{2+i}=\frac{1}{2+i}\cdot \frac{2-i}{2-i}=\frac{2-i}{5}=\frac{2}{5}-\frac{1}{5}i.$$Now we have two nonreal roots of $f$. Since $f$ has real coefficients, the conjugates of its roots are also roots. Therefore, the four roots of $f$ are $2\pm i$ and $\frac{2}{5}\pm \frac{1}{5}i$.

The monic quadratic whose roots are $2\pm i$ is $(x-2-i)(x-2+i)=(x-2{)}^2-i^2=x^2-4x+5$.

The monic quadratic whose roots are $\frac{2}{5}\pm \frac{1}{5}i$ is $\left(x-\frac{2}{5}-\frac{1}{5}i\right)\left(x-\frac{2}{5}+\frac{1}{5}i\right)={\left(x-\frac{2}{5}\right)}^2-{\left(\frac{1}{5}i\right)}^2=x^2-\frac{4}{5}x+\frac{1}{5}$.

Therefore, $$\begin{array}{rl}f(x)& =a(x^2-4x+5)\left(x^2-\frac{4}{5}x+\frac{1}{5}\right)\\ & =a\left(x^4-\frac{24}{5}{x}^3+\frac{42}{5}{x}^2-\frac{24}{5}x+1\right),\end{array}$$so $a,b,c$ are in the ratio $1:-\frac{24}{5}:\frac{42}{5}$. Since $a,b,c$ are integers whose greatest common divisor is $1$, we have $(a,b,c)=(5,-24,42)$ or $(-5,24,-42)$. In either case, $|c|=\boxed{42}$.