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

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    $\begingroup$ "i have checked for most primes" --- that must have taken a while! $\endgroup$ – John Hughes Oct 23 '17 at 18:20
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    $\begingroup$ "If $p$ is prime then there exists integers $a,b$ such that $$p=a+b \text{ and } ab=n \text{ and } \frac{n}{n_1} + \frac{n}{n_2} = (a+b) \pm 1$$ where $n=n_1 n_2$." Please ensure that this is indeed the statement you wish to prove. $\endgroup$ – Andrew Tawfeek Oct 23 '17 at 18:47
  • $\begingroup$ Sorry just meant to ensure I didn't just check for the first few primes in fact why even apologize? Andrew Tawfeek yes that is correct $\endgroup$ – argamon Oct 23 '17 at 19:02
  • $\begingroup$ John Hughes yes it took 48 hours on a laptop to check primes up to 1 million $\endgroup$ – argamon Oct 23 '17 at 19:08
  • $\begingroup$ Updated the question so it is more clear what is asking, please correct if needed. $\endgroup$ – Sil Oct 23 '17 at 19:35

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 $a+b=c+d-1$

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    $\begingroup$ That's a great answer thank you Peter $\endgroup$ – argamon Oct 24 '17 at 15:16
  • $\begingroup$ This shows that there must be one solution so my next quest is how do we know whether or not there is a solution of a number is congruent 0 mod 3 $\endgroup$ – argamon Oct 26 '17 at 19:00
  • $\begingroup$ @argamon Seems for $p\equiv 0\mod 3$, there are infinite many counterexamples. $\endgroup$ – Peter Oct 27 '17 at 22:04
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    $\begingroup$ Haha you should be $\endgroup$ – argamon Oct 28 '17 at 20:39
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    $\begingroup$ Haha ok thanks Peter one thing that mivht be intersting is that ive found other numbers like 144 and 180 where there are 3 consecutive pairs so like a,b =c+d-1=e+f-2 and ab=cd=ef $\endgroup$ – argamon Oct 30 '17 at 23:48

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