TITLE: Prime number written as a sum
QUESTION [5 upvotes]: I have a question related to prime numbers which has been tough for me to get forward with. I wonder whether every prime number can be written as the sum of two numbers such that the product of the two numbers yields another number which can be factored into an additional two pairs whose sum happens to differ by plus or minus $1$ from the original pair.
In other words, is following true:

For every prime $p$ there exist positive integers $a,b,c,d$ satisfying $ab=cd$ such that $p=a+b$ and $c+d=p\pm1$.

So for example $7= 4+3$ and $3\cdot4=12$. Then $12$ can be factored as $6\cdot2$ and $6+2 =8$ which differs by $1$ from the $4+3$. I have checked for most primes but is this trivial ? If not is there a standard method to start with trying to prove this? I am sorry for not using math format but just curious to get some feedback.  Thank you

REPLY [6 votes]: If $p\equiv 2\mod 3$, then the choice $$a=\frac{2p+2}{3}$$ $$b=\frac{p-2}{3}$$ $$c=\frac{p+1}{3}$$ $$d=\frac{2p-4}{3}$$ does the job ; we have $ab=cd$ and $a+b=c+d+1$. In the case $p\equiv 1\mod 3$, choose $$a=\frac{2p-2}{3}$$ $$b=\frac{p+2}{3}$$ $$c=\frac{p-1}{3}$$ $$d=\frac{2p+4}{3}$$ Then, we have $ab=cd$ and $a+b=c+d-1$