   SEARCH HOME Math Central Quandaries & Queries  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     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.