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Counting the number n of positive (x,y) = (p,q), p and q primes, solutions to third degree Thue equations equal a prime r, and p,q,r <1000 i get with the Pari gp built-in functions thueinit() and thue(), in a non exhaustive search:

n=9 for $3x^{3} -3x^{2}y +y^{3} =r $. They are, written in the form $(p,q,r)$ :

$(3,5,71);(2,5,89);(5,7,193);(5,2,233);(2,7,283);(7,5,419);(7,2,743);(7,11,743);(5,11,881)$

Any particular third degree Thue equation equal a prime with a number n of positive prime solutions less than 10^3; n >9 ?

I would be very surprised if anybody could find a n >13

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Try meta.math.stackexchange.com/questions/3399/… and the various threads linked thereunto. –  Gerry Myerson Jan 22 '13 at 10:57
    
Let me rephrase your question. I think you are asking whether there is a cubic form $f(x,y)$ with integer coefficients such that there are more than $9$ triples $(p,q,r)$ with all entries between $0$ and $1000$ such that $p$, $q$, and $r=f(p,q)$ are all prime. Is that your question? –  Gerry Myerson Jan 22 '13 at 22:55
    
It would be my question if you added the condition for the cubic polynomial in two variables to be homogenous: Sum (a(i)*x^i*y^(3-i)) = r 2<=p,q,r<1000; p,q,r primes. –  user55514 Jan 22 '13 at 23:50
    
Note that I wrote "cubic form," not "cubic polynomial." Part of the definition of "form" is "homogeneous". Anyway, how do you like my answer? –  Gerry Myerson Jan 23 '13 at 1:04
    
I liked your answer. A correct answer is always a good response - i could not avoid the redundancy. But you limited yourself to just giving the least answer you could (n =10) as i remember i used to do when i was at high school at Orthez in France in a non apropriate place for me; but i did not know it then; and i was (in he deep) "scared" by what the teachers and my father demanded from me at school, and always limited myself to the least effort in a way. I answered all they asked, but stopped there. My brain -they changed it and i did not know it- is a bit better now ; awakening a bit. –  user55514 Jan 23 '13 at 10:28

2 Answers 2

This smacks somewhat of numerology, but I'll bite. Hows about $p^3-4p^2q+ p q^2+7q^3=r$, for $$ \begin{array}{c} (p,q,r)=(2,3,167), (2,5,853), (3,2,23), (3,5,797), (5,3,59), \\ (7,3,7), (11,2,463), (11,3,167), (11,5,61), (11,7,883), (13,2,953), \\ (13,5,17), (13,7,503), (17,5,433), (19,7,83), (23,7,883), (23,11,991), \\ (29,11,211), (29,13,937), (41,17, 853), (43,17,593), (47,19,919). \end{array} $$ This was found with a short search. One would expect to find examples with many more such values, if one was so inclined.....

P.S. Hi Gerry!

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Not numerology. The exact value of the primes does not matter. Just trying to calibrate how many of those solutions might exist. I was just wrong thinking of a limit of 20 solutions for a third degree equation and probably less if the degree increases,in the same range of primes, but again i might be wrong. –  user55514 Jan 23 '13 at 10:13
1  
Hi, Mike --- well done! If you hunt around a bit, you'll probably find some other questions on this site that could do with a bit of your expertise. –  Gerry Myerson Jan 23 '13 at 12:10

$p^3-p^2q+q^3=r$ is satisfied by the $10$ prime triples $$(p,q,r)=(2,3,23),(2,5,113),(3,2,17),(3,5,107),(3,7,307),(5,2,83),(5,7,293),(7,3,223),(7,5,223),(11,7,827),$$ all with positive entries below $1000$.

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