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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 a2 - ab - 2b2? 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 prime-generating 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 x2 - xy + y2. 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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