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| Math Central | Quandaries & Queries |
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Dear Sirs, Is it possible for (a^2 - a.b + b^2) to be divisible by a prime p where b > a > p and a, b, and p are relatively prime? I let a = n.p + x and b = m.p + y (where both x and y are < p) and after dividing throughout by p one gets to decide whether (x^2 - x.y + y^2) ==0 mod(p) ?! It seems unlikely unless p = (x^2 - x.y + y^2)? (Note:- All letters represent positive integers.) I'd be grateful for your thoughts and comments Many Thanks, Andrew |
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!
Let b=11 and a=10. then a2 - ab + b2 = 100 - 110 + 121 = 111 = 3*37.
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:
http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html
Chris
Andrew,
Try a few examples using p = 3, a == 1 (mod 3) and b == 2 (mod 3) suitably chosen
to satisfy your other constraints. That should show you what's going on.
Victoria
Andrew wrote back
Question from Andrew:
Dear Chris and Victoria,
Please could you give me any other solution for any prime p > = 5
where (a, b, 2) = 1?Many Thanks Again!
Andrew
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
there exists an integer x for which x2 - x + 1 can be divided by the prime p if and only if p = 3 or p-1 is divisible by 3.
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.
(In case you do not understand my notation, for the mod 7 example, you can take x = 3, 10, 17, ..., 3 + 7k, ..., and y to be 1, 8, 15, ... 1 + 7n. Neither size nor parity is relevant here: x2 - xy + y2 will be divisible by 7.)
Chris
Andrew,
For p=7 try a=15 and b = 31. Then,
a2 - ab +b2
== 12 - 1x3 + 32
==1 - 3 + 2 == 0 (mod 7).
Victoria
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