



 
Andrew, There usually isn't much one can say about the integer values of polynomials over the integers that fail to factor in an obvious way  usually their values are sometimes prime, sometimes not. With a and b relatively prime, there would be no obvious way of predicting the size of the prime factors. For your example, the first values for a and b that I tried happened to satisfy all your criteria! Could you be thinking of polynomials that factor, such as a^{2}  ab  2b^{2}? Since this factors into (a  2b)(a + b), the values it takes on will be composite integers as a and b run through the positive integers and a is larger than 2b+1. On the other hand, a theorem of Goldbach (1752) tells us that no polynomial over the integers , even with several variables, will produce only prime numbers. You might be interested in the article on primegenerating polynomials on the Mathworld web page: Chris
Andrew, Try a few examples using p = 3, a == 1 (mod 3) and b == 2 (mod 3) suitably chosen Victoria Andrew wrote back
Andrew, What one really needs in discussing divisibility of quadratics by primes is a knowledge of QUADRATIC RECIPROCITY (which one studies in an elementary number theory course). First, you don't lose generality by letting y = 1 in your quadratic x^{2}  xy + y^{2}. I don't guarantee it, but off the top of my head I believe the result here is that
Thus for p = 3 you can let x be 2 (mod 3); for p = 7 you can let x be 3 or 5 (mod 7); for p = 13 you can let x = 4 or 10 (mod 13); etc. Chris Andrew, For p=7 try a=15 and b = 31. Then, Victoria
 


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