let $x=a-2b+c,y=b-2c+a$ we have $A=2x^4+2y^4+2(x+y)^4=2[2(x^2+xy+y^2)^2]=(2x^2+2xy+2y^2)^2$ and for $B$, $B=d(d+1)(d+2)(d+3)+1=(d^2+3d)(d^2+3d+2)+1=(d^2+3d+1)^2$ $(\sqrt{A}+1)^2+B\equiv (2x^2+2xy+2y^2+1)^2+(d^2+3d+1)^2 mod 4 \equiv 1+1 \equiv 2 mod 4$ Which is never be a perfect square since $2$ is not a quadratic residue in modulo $4$

You can also use mod 3 . In this problem, you should just know some lemmas in algebra.

Let $x=a-2b+c$ and $y=b-2c+a$ $A=2x^4+2y^4+2(x+y)^4=4[x^4+y^4+2xy(x^2+y^2)+6x^2y^2]$ $B=(d^2+3d+1)^2$ Suppose that there exist an integer $k$ such that $(\sqrt{A}+1)^2+(d^2+3d+1)^2=k^2$ $\sqrt{A}+1, d^2+3d+1, k$ is a Pythagorean triple. Pythagorean triples are in the form $u^2-v^2, 2uv$ and $u^2+v^2$. $d(d+3)+1 \equiv 1 mod 2$ which implies $d(d+3)+1$=$u^2-v^2$ and $\sqrt{A}+1=2uv$ $=>$ $A \equiv 1 mod 2$. Contradiction. Hence $(\sqrt{A}+1)^2+B$ is not a perfect square

Correct me if I am wrong (I think we can solve iit for all integer A) At first assume for contradiction After expanding $(\sqrt A+1)^2+B$ we get that $A+B+1+2\sqrt A$ must be perfect square(integer) $\implies \sqrt A$ must be rational and we know that A is integer then $\sqrt A$ must be integer the rest is same as @2 above

Accually... its much easier by using some lemmas.... lemma 1 : a^4 + b^4 + c^4 - 2( (bc)^2 + (ab)^2 + (ac)^2) = (a+b+c)(a-b+c)(a+b-c)(a-b-c) and lemma 2 : if a+b+c=0 then (ab)^2 + (bc)^2 + (ca)^2= (ab+bc+ca)^2 Now let a-2b+c = x, b-2c+a=y and c-2a+b=z. Now using lemmas 1, 2 in A we get sqrt (A) = 2(xy+yz+zx) and after that it is done mod 4

$A$ is even so $(\sqrt{A}+1)^2 \equiv 1(mod 4)$ $B \equiv 1(mod 4)$ so $(\sqrt{A}+1)^2+B \equiv 2(mod 4)$