Proving $a(a+d)(a+2d)(a+3d)+d^4$, with integer $a$ and $d$, is a perfect square Question:

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer.

Solution given everywhere
Let the terms be $a-3d$, $a-d$, $a+d$ and $a+3d$ ( with common difference $2d$).
$(a-3d)(a-d)(a+d)(a+3d) + (2d)^4$ has to be proved a perfect square
which can be further written as $(a^2-5d^2)^2$.
And now they have stopped! They have concluded $a^2-5d^2$ as an integer just by saying that all the terms are integer!
But I think the solution is incomplete. $a$ and $d$ are not given integers. So we cannot surely say that $a^2 - 5d^2$ is an integer. For completion, we have to take cases that whether $2d$ is odd or even. But that will make the answer a bit lengthy.
My approach
Since the answer will be a bit lengthy in previous case, so I took the terms as $a$, $a+d$, $a+2d$, and $a+3d$ with common difference $d$ so that I can surely say that a and d are integers.
So we have to prove $a(a+d)(a+2d)(a+3d)+d^4$ as the square of an integer
which can be further simplified as:
$$a^4 + d^4 + 6a^3 d + 6a d^3 + 11a^2 d^2 \tag{$\star$}$$
I am stuck at this step.

How can I do further to make $(\star)$ a perfect square?

 A: Simply note that
$$ a(a+d)(a+2d)(a+3d) + d^4 \\
=  a^4 + 6 a^3 d + 11 a^2 d^2 + 6 a d^3 + d^4\\
=   (a^2 + 3 a d + d^2)^2 $$
A: Note that there is a simple linear relation between your $a, d$ and the given solution's $a,d$. Therefore, you can find that relation, and substitute that into the $(a^2-5d^2)^2$ of the given solution, to obtain a perfect square in terms of your $a$ and $d$. But note that it will involve a denominator of 2. Therefore, you need to analyze the parity of $a$ and $d$.
And as for the given solution, you can also determine that $a$ and $d$ are of the form $p/2$ and $q/2$, where $p,q$ have the same parity. Therefore you can safely conclude that $(a^2-5d^2)$ is an integer.

So let's tackle the expression $a^4 + d^4 + 6a^3d + 6d^3a + 11a^2d^2$. Note that every term is the product of 4 variables (e.g. $a^3d$ is 3$a$'s and one $d$), so we say it is homogeneous. Appearantly, the perfect square should also be homogeneous (can you see why?). Also, it is symmetric in $a$ and $d$. So the final result should be $$(?_1a^2 + ?_2ad + ?_1 d^2)$$ where $?_1$ and $?_2$ are to be determined.
Now look at the coeffecient of $a^4$ and $d^4$ we see that $?_1 = 1$. So $$(a^2 + ?_2 ad + d^2)^2 = a^4 + d^4 + 6a^3d + 6d^3a + 11a^2d^2.$$ Now compare the coeffecient of $a^2d^2$ on both sides: $2 + ?_2^2 = 11$. This means that $?_2 = \pm3$. A moment of thought reveals that it should be 3. And bravo: $$a^4 + d^4 + 6a^3d + 6d^3a + 11a^2d^2 = (a^2 + 3ad + d^2)^2.$$
