Math CentralQuandaries & Queries


Dear Victoria,

I hope you are 'just fine'!

Another way of looking at the 'alternating parity polynomial', again based on Fermat's Little Theorem,
is to substitute (a - b) for x in x^(p-1) - 1 as this is always divisible by any prime p. So, if one removes
the "- 1", there is always a remainder of (1/p)! (I took up your challenge!)

Again, in an effort to exclude primes, I introduced a third term to go symmetrically with the second
added variable. This gives:-

a^2 + a.b + b^2 == c^2 - c.a + a^2 == c^2 - c.b + b^2 mod(p)

Interestingly, if I choose any pair in the above, I get a similar answer to the one you gave me.
How does it work for the three polynomials taken together?
Do they "exclude" primes?! (I'm looking for a mathematical padlock!)

If I summate them I get 2.(a^2 + b^2 + c^2) + a.b - c.a - c.b == 0 mod(p)
and this doesn't appear to have (a + b - c) as a factor?

In this question, I forgot to mention the parameters are as before1 Perhaps, to avoid trivial answers,
one of the variables (a, b, c) should be assumed 'even'. Say, b is 'even'.

If the congruencies work, say, for a == b == c mod(3), are there any other odd primes > 3 which are not excluded?

i.e. Might all primes p > 3 be excluded but not p = 3 itself? (Can you give an example?!)

Thanks Again, I really appreciate your time and thoughts!

Take Care,

All the Best,


PS Is "Victoria West" a nom-de-plume?! (I can't find you at UoR but I found Chris!)


I am no expert in number theory. Although we can answer specific questions, we are not much help with vague speculation. It looks as if you have an interest in the type of problems that arise in elementary number theory classes, so why not get hold of a textbook for such a course and work your way through the book. My guess is that your questions can be reduced to problems that can be solved with modular arithmetic combined with some knowledge of quadratic reciprocity. This is a field that benefits from a systematic approach (plus a lot of experimentation, either the old fashioned way with pencil and paper, or the modern way with a computer).




I worked with Chris and Harley at the University of Regina a long time ago.

It doesn't look to me as if adding a third term will help much. See what you make
of this (and check that it is correct!).

If you look at c2 - c.a + a2 == c2 - c.b + b2 (mod p), then after rearranging
you get c(b-a) + (a2 - b2) = (a-b)(a+b - c) == 0 (mod p). Therefore p divides
(a-b), so that either a is congruent to b (mod p), or p divides (a+b-c).

If you look at a2 + a.b + b2 == c2 - c.a + a2, then after rearranging you
get a(b+c) + (b2 - c2) = (b+c)(a+b-c) == (mod p). Therefore either
b is congruent to -c (mod p) or p divides (a+b-c).

In order for the (triple) congruence to hold, one of each pair of conditions must
hold. That is, either (a+b-c) == 0 (mod p), or a == b == -c (mod p).

In the second case, the sum of the three terms becomes 9a2, which is
divisible by p only if p = 3 or a == 0 (mod p).

In the first case you have a+b == c (mod p), so the sum of three terms
2(a2 + b2 + (a+b)2) +ab - a2 - ab - ab - b2
== a2 -ab + b2 + 2(a+b)2
== 3a2 + 3ab + 3b2
== 3 (a2 + ab + b2) (mod p).
This will be congruent to 0 (mod p) if and only if p = 3 or
a2 + ab + b2 == 0 (mod p). We've handled the last of these before.

Good luck!


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