Solution:
First we prove the following Lemma
Lemma: If $x$, $y$, $z$ are real numbers such that $x + y + z = 0$, then $2\left(x^{4} + y^{4} + z^{4}\right) = \left(x^{2} + y^{2} + z^{2}\right)^{2}$.
Proof of the Lemma:
$$\begin{aligned} x^{4} + y^{4} + z^{4} & = x^{2} x^{2} + y^{2} y^{2} + z^{2} z^{2} = x^{2}(y+z)^{2} + y^{2}(z+x)^{2} + z^{2}(x+y)^{2} \\ & = 2\left(x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2}\right) + 2 x y z(x + y + z) \\ & = \left(x^{2} + y^{2} + z^{2}\right)^{2} - x^{4} - y^{4} - z^{4} \end{aligned}$$
and the claim follows.

Now back to our problem notice that $(a-2b+c) + (b-2c+a) + (c-2a+b) = 0$, thus according to the lemma it holds
$$\begin{aligned} A & = 2(a-2b+c)^{4} + 2(b-2c+a)^{4} + 2(c-2a+b)^{4} \\ & = \left[(a-2b+c)^{2} + (b-2c+a)^{2} + (c-2a+b)^{2}\right] = \left[6\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right)\right]^{2} \end{aligned}$$
Since $a^{2} + b^{2} + c^{2} \geq ab + bc + ca$ we have that
$$\sqrt{A} + 1 = 6\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) + 1$$
In addition, it is easy to check that
$$B = d(d+1)(d+2)(d+3) = \left(d^{2} + 3d + 1\right)^{2}$$
Let us set $6\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) + 1 = m$, $d^{2} + 3d + 1 = n$. We need to prove that the number $(\sqrt{A} + 1)^{2} + B = m^{2} + n^{2}$ is not a perfect square.
Since both $m$, $n$ are odd integers, both $m^{2}$, $n^{2}$ are integers of the form $4k + 1$, so the number $m^{2} + n^{2}$ is an integer of the form $4k + 2$. But it is well known that all perfect squares are of the form $4k$ or $4k + 1$, and we are done.