Prime or composite?

Letting $c=a$ and $b=d=-a$ the condition is met and $a+b+c+d=0$ is neither composite nor prime. So in order to clarify the problem, I would advise to change the domain of $a, b, c, d$ into the positive integers. With this assumption the answer can be proven in the following way:

From $a^2+b^2+c^2+d^2=2(a^2+b^2)$ we have

$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd=2(a^2+b^2+ab+ac+ad+bc+bd+cd$ ,

hence $2 | (a+b+c+d)^2$ .

Since an even square number can only be the square of an even integer, it follows that $2 | a+b+c+d$ . In view of $a+b+c+d \geq 4$ , $2$ is a proper divisor of $a+b+c+d$ . Thus the conclusion follows.

Assume $a + b + c + d$ can be not a composite number i.e. a prime number. Since $a + b + c + d$ is prime, it must be odd, because $a + b + c + d = 2$ has no solutions in positive integers. By checking combinations of parities, we see that either exactly one or exactly three of $a, b, c, d$ have to be odd in order for the sum to be odd.

Now, let's examine $a^2 + b^2 = c^2 + d^2.$ Suppose exactly one of $a, b, c, d$ were odd. Then, no matter which one of those numbers was odd, we must have $\text{even}^2 + \text{even}^2 = \text{even}^2 + \text{odd}^2,$ which simplifies into $\text{even} + \text{even} = \text{even} + \text{odd},$ since a number retains its parity when it is squared. This clearly cannot occur, so there can't be exactly one odd number among $a, b, c, d.$

Suppose exactly three of $a, b, c, d$ were odd. This means that exactly one of them is even. Then, no matter which one of those numbers was even, we must have $\text{odd}^2 + \text{odd}^2 = \text{odd}^2 + \text{even}^2,$ which simplifies into $\text{odd} + \text{odd} = \text{odd} + \text{even}.$ Since an odd plus an odd is even, this cannot occur, so there can't be exactly three odd numbers among $a, b, c, d.$

Thus, we have reached a contradiction, and the value of $a + b + c + d$ must be a composite value. $\blacksquare$

If A and B are both odd or both even, then the left side of the equation is an even number, and right side must also be even. (A+B) will be even since the sum of two odd numbers is an even number, and the sum of two even numbers is an even number. (C+D) will also be even because the only way to make the right-hand side equal is to make C and D either both even or both odd, which will result in an even sum either way.

If one and only one of A and B is odd, then the left-hand side will be odd. The right-hand side will also be odd, and the only way to obtain an odd sum from two integers is if one of them is even and one of them is odd. Therefore, one and only one of C and D is odd. When you take the sum (A+B+C+D), it will turn out something like this:

O + E + O + E

Which will always be an even number. Now, the only case we have to consider is A+B+C+D = 2, but it's easily dealt with since the minimum value of A+B+C+D is 4.

So, since A+B+C+D will always be even but will never be 2, it will always be composite.

Here is an 'elementary' solution involving parity. Consider the following two partitioning cases:

Case 1: Suppose both sides are even. Then it follows that a and b have the same parity (i.e both are odd or both are even) and the same applies to c and d [to see this consider what would happen if they had the same parity]. This means that a+b and c+d are both odd (as 'odd+even = odd') so a+b+c+d is even but as they are all positive integers $\ a+b+c+d \geq 4$ so a+b+c+d is composite

Case 2: Suppose now that both sides are odd. This time it follows that (a,b) and (c,d) are pairs of positive integers with opposite parity so by similiar reasoning a+b and c+d are both odd so a+b+c+d is even, again implying that it's value is composite. And we are done