Math CentralQuandaries & Queries


Dear Chris and Victoria,

Your answers have helped me with my more complex equation which has a factor (on re-arranging it) of
(a^2 + a.b + b^2).

As I mentioned to you, I am trying to exclude primes rather than generate them! How effective are the two pairs of polynomials, taken together, at "excluding" any prime p.

1. (a^2 - a.b + b^2) and (a^2 + a.b + b^2) Where b > a > p and (a, b, p) = 1

2. (c^2 - c.a + a^2) and (c^2 - c.b + b^2) Where c > b > a > p and (a, b, c, p) = 1

In the second case, I have introduced an extra variable in the hope, Chris and Victoria, that it will
preclude letting "y = 1", so-to- speak!

From your previous answers, it would seem most unlikely (impossible?!) that a prime p would divide both polynomials in the second case (2) but it may still be possible for a prime p to divide both polynomials in the first case (1)?

I really look forward to receiving your answers! By the way, is it your mathematical intuition that leads you both so quickly to your solutions or do you take awhile? It seems to take me ages!

If you feel I've asked two questions instead of one, please would you answer the second case (a, b, c) preferentially!

Many Thanks,



I don't know if it is intuition or just experience. One the one hand, mathematics is often much more of an art than it is anything else. On the other hand, after enough practice looking at problems of a certain type, one develops a feeling for what kinds of arguments might lead to solutions.

If p is prime and divides both (a2 - a.b + b2) and (a2 + a.b + b2), then (a2 - a.b + b2) == (a2 + a.b + b2) (mod p). Therefore, 2ab == 0 (mod p),from which it follows that p|a, or p|b, or p=2. Pushing this a little further, if p is an odd prime and p | a and also p | (a2 - a.b + b2), then p|b. Thus there are no solutions with (a,b,p)=1 when p is an odd prime. When p = 2, a similar analysis shows that a and b must also b even. Hence there are no solutions of the type you want.

In a similar way, the second case leads to a2 - ca == b2 - cb (mod p), which leads to:
            a2 - b2 + c(a-b) = (a+b)(a-b) + c(a-b) = (a-b)(a+b+c) == 0 (mod p).
Thus p | a-b or p | a+b+c. It looks like the introduction of the second variable does not help restrict things farther. For example, take p=3, a = 7, b = 10, and c = 11. Then c2 - c.a + a2 == 1-2+1 == 0 (mod 3), and similarly for c2 - c.b + b2. (All I have done is taken a == b (mod 3).)

Good luck!


Dear Victoria,

Thanks so much! Should I be fortunate enough to meet you one fine day, I'll have to wear two pairs of dark glasses
rather than one! (I hope you don't mind the compliment?!)

I need to go carefully through all that you've said and see where I get to.

I've been looking at the polynomial [ap - 1 + ap - 2 + ap - 3 + .... + a + 1] == 0 mod(p) and trying to apply your principles to it, as well as its counterpart where the parity alternates between terms and hoping to discover that p will not divide?!

So much appreciate your help and good luck wishes!

I hope all goes much better for you than you could ever dream possible in 2009!

Take Care,

All the Best,




Although you didn't ask, here are some thoughts on your problem. This is off the top of my head. If it isn't correct, maybe you can correct it.

The prime p will divide ap - 1 + ap - 2 + ap - 3 + .... + a + 1 if and only if a is congruent to 1 (mod p). Here is how to see that. If p divides a, then the sum is clearly congruent to 1 (mod p) since all but one term is zero (mod p). If a is congruent to 1 mod p, then the sum is p == 0 (mod p). Otherwise, there is a two step argument. First, ap - 2 + ap - 3 + .... + a + 1 = (ap-1-1) / (a-1). Since a isn't congruent to 1 (mod p), a-1 isn't congruent to 0 (mod p), and so p divides (ap-1-1) / (a-1) if and only if it divides ap-1 -1. The latter statement is exactly Fermat's (Little) Theorem. Therefore, your original sum ap-1 + ap-2 + ap-3 + .... + a + 1 == ap-1 + 0 == 1 (mod p), again using Fermat's Theorem.

When the signs alternate, the sum is congruent to 0 (mod p) if and only if 1 + a2 + a4 + ... + ap-1 == a + a3 + ... + ap-1 (mod p). This is not true if a is congruent to 0 (mod p). Otherwise, suppose for a moment that a2 is not congruent to 1 (mod p). Then the LHS is [(a2)(p+1)/2 - 1] / (a2 -1) = [ap+1 - 1] / (a2 - 1) and similarly the RHS is a[ap - 1] / (a2 - 1). Since a2 - 1 is not zero (mod p), these two quantities are congruent if and only if ap+1 - 1 == ap - a (mod p). But, by Fermat's Theorem ap - a == 0 (mod p) and ap+1 - 1 == a2 - 1 (mod p). By assumption, this quantity is not zero (modulo p). Thus the only time one could hope that 1 + a2 + a4 + ... + ap-1 == a + a3 + ... + ap-1 (mod p) is when a2 == 1 (mod p). In that case, the LHS is 1 (mod p) and the RHS is (p-1)/2. The only prime p for which 1 == (p-1)/2 (mod p) is p=3. In summary, when the signs alternate, the sum is congruent to 0 (mod p) if and only if p=3 and p does not divide a (which means that a2 == 1 (mod p)).

Best of luck.


Hi Victoria,

Could you confirm (if I'm reading you right!), that in the case of the alternating parity polynomial, no prime will divide
into it except prime p = 3 given (a, p) = 1?

i.e. a^36 - a^35 + a^34 - a^33 + ..... + a^2 - a + 1 =/= 0 mod (37) ?! (Yes, you guessed?! "37" is my 'lucky number'!)

My problem is in two parts and, if the above is true, then it may help greatly with "Part II"!

Thanks so much, again.

Take Care,



Yes, you did read what was written correctly. You should carefully check the argument because it was done off the top of my head and may not be correct. If it is wrong, then perhaps there are enough decent ideas to lead you to your own solution.

Good luck!

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