For some natural numbers $a, b, c, d$ if $a\cdot b = c \cdot d$ is it possible that the sum $a + b + c + d$ is prime number, or in other words it doesn't have divisors rather then $1$ and itself.
What I tried
I think that the answer is not, here is my try
Case 1: All numbers $a, b, c, d$ are even, the sum will be even too, so in this case $a+b+c+d$ cannot be prime number
Case 2: One number of $(a, b)$ is even and one is odd, let's say $a$ is even, and $b$ odd. Let it hold for $(c, d)$ too, $c$ is even, $d$ is odd. Obviously $b + c$ is going to be an even number, because two odd numbers sum up to an even number. When we add $a$ and $c$ to this sum, we will end with an even number, and not prime number.
Case 3: Exactly one of the $4$ numbers $a, b, c, d$ is odd. For simplicity let it be $a$. It is obvious that the sum $a+b+c+d$ is going to be odd number.
I'm stuck on case 3, can you give me some hints how to continue and finish the proof.