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

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cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.0Similar to mukushla's previous post.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You can I think show that pq =2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[pq + p + q + 2 = p^2\]\[q=\frac{p^2p2}{p+1}=\frac{p^2+p2p2}{p+1}=p2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[pq+p+q+2=p^2\]\[p(q+1)+q+2=p^2\]\[(p+1)(q+1)+1=(p+1)(p1)+1\]\[q+2=p\]\(p\) 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
 3 years ago
Best ResponseYou've already chosen the best response.0i was lookin for it ... thank u herp derp...its open

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Impossible to solve!!!?!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Open 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.
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