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Given the Diophantine equation$$ax^2+bxy+cy^2+dx+ey+f=0$$ if the coefficients $(a,b,c,d,e,f)$ are chosen among all the prime numbers, we have infinite equations. Is it possible to prove that the solutions of the infinite equations are infinite and countable?

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closed as off topic by Gerry Myerson, Andreas Blass, Felipe Voloch, Emil Jeřábek, Pietro Majer Feb 17 '12 at 12:45

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If $a$ and $c$ are large enough compared to the other coefficients then the left side will be positive for all $x,y$ so there will be no solutions. Pietro has shown that if you take the full set of equations (or even a small chunk) you get infinitely many solutions. Simple cardinality considerations show it's impossible to have an uncountable infinity of solutions, even taking all the equations. I see no question at a research level here, so I vote to close. –  Gerry Myerson Feb 16 '12 at 22:24
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Could you tell me why is this interesting? –  Woett Feb 16 '12 at 22:37
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1 Answer 1

up vote 1 down vote accepted

Yes, if e.g. $a=e=2,\, b=5,\,c=d=3,$ and $f$ varies among all primes, that equation has the solution $x=-y=f$, which already makes infinitely many (distinct) solutions.

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(quite a stupid answer, but an answer) –  Pietro Majer Feb 16 '12 at 21:33
    
It's a start... –  Spice the Bird Feb 16 '12 at 22:11
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actually I didn't mean to give a start –  Pietro Majer Feb 17 '12 at 0:04
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