Solve in prime numbers pq + p + q + 2 = p^2.

Question

cwrw238 · Answer

Similar to mukushla's previous post.

cwrw238 · Answer

both p and q are prime

anonymous · Answer

You can I think show that p-q =2

anonymous · Answer

Here is a sequence for up to 100th prime of (p,q)

{{5, 3}, {7, 5}, {13, 11}, {19, 17}, {31, 29}, {43, 41}, {61, 
  59}, {73, 71}, {103, 101}, {109, 107}, {139, 137}, {151, 149}, {181,
   179}, {193, 191}, {199, 197}, {229, 227}, {241, 239}, {271, 
  269}, {283, 281}, {313, 311}, {349, 347}, {421, 419}, {433, 
  431}, {463, 461}, {523, 521}}

anonymous · Answer

$$pq + p + q + 2 = p^2$$$$q=\frac{p^2-p-2}{p+1}=\frac{p^2+p-2p-2}{p+1}=p-2$$

anonymous · Answer

$$pq+p+q+2=p^2$$$$p(q+1)+q+2=p^2$$$$(p+1)(q+1)+1=(p+1)(p-1)+1$$$$q+2=p$$$p$ and $q$ are twin primes... But it's an open conjecture on whether there are an infinite number of twin primes... http://mathworld.wolfram.com/TwinPrimeConjecture.html

anonymous · Answer

i was lookin for it ... thank u herp derp...its open

anonymous · Answer

Impossible to solve!!!?!

anonymous · Answer

Open Questions of primes http://primes.utm.edu/notes/conjectures/ i have a nice article about open questions...i cant find it now..i'll upload it later.

cwrw238 · Answer

thanks guys