## cwrw238 3 years ago Solve in prime numbers pq + p + q + 2 = p^2.

1. cwrw238

Similar to mukushla's previous post.

2. cwrw238

both p and q are prime

3. eliassaab

You can I think show that p-q =2

4. eliassaab

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}}

5. mukushla

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

6. Herp_Derp

$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

7. mukushla

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

8. Herp_Derp

Impossible to solve!!!?!

9. mukushla

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.

10. cwrw238

thanks guys