Note: I don't know how to program a computer to check for me, hence my question. Thanks in advance.
Building on the concept explored in
Sets of Prime and Composite Numbers
Conjecture: For every pair of twin primes $a$ and $b$, performing the arithmetic operation of:
$(a+b)+(a ⋅b)= c$
[Example: $(3+5)+(3⋅5)=23$]
yields either a composite number part of a Twin-Composite pair or a prime only within a Prime-Composite pair.
[Example: Twin Composites, Example: 119,121 (composite followed by a composite)]
[Example: Prime-Composite, Example: $23,25$ (prime followed by a composite)]
First Ten Examples from Manual Calculation:
$a = 3$ and $b = 5$ ; yields $23$ ; $23$ is prime, $25$ is composite
$a = 5$ and $b = 7$ ; yields $47$ ; $47$ is prime, $49$ is composite
$a = 11$ and $b = 13$ ; yields $167$ ; $167$ is prime, $169$ is composite
$a = 17$ and $b = 19$ ; yields $359$ ; $359$ is prime, $361$ is composite
$a = 29$ and $b = 31$ ; yields $959$ ; $959$ is composite, $961$ is composite
$a = 41$ and $b = 43$ ; yields $1847$ ; $1847$ is prime, $1849$ is composite
$a = 59$ and $b = 61$ ; yields $3719$ ; $3719$ is prime, $3721$ is composite
$a = 71$ and $b = 73$ ; yields $5327 $ ; $5327$ is composite, $5329$ is composite
$a = 101$ and $b = 103$ ; yields $10607$ ; $10607$ is prime, $10609$ is composite
$a = 107$ and $b = 109$ ; yields $11879$ ; $11879$ is composite, $11881$ is composite
Conclusion: $c$ is either part of a twin-composite pair or a prime-composite pair.
Best Answer
As @DanielFischer pointed out, the conjecture is true:
$a=b-2$
$\Rightarrow c=(a+b)+(a*b)$
$=(b-2+b)+((b-2)*b)$
$=b^2-2$.
Thus, $c+2=b^2$ is always composite.