Math Forum Discussions

Christian Bau

Posts: 720
Registered: 12/6/04

Re: JSH: Basic result, test REVISED
Posted: Jan 20, 2004 6:27 PM

Plain Text

In article <pmo0e1-vil.ln1@hudson.theathertons>,
Mark Atherton <n_z_w_nguregba@lnubb.pb.hx> wrote:

> James Dolan wrote:
> > in article <3c65f87.0401201027.31e1abd6@posting.google.com>,
> > james harris <jstevh@msn.com> wrote:
> >
> > |One hallmark of new research is the ability to answer questions that
> > |others might not have even considered. I've been playing with a
> > |result that I've mentioned before, but became curious enough to post
> > |it again, and see if I'm wrong in assuming that what's easy for me to
> > |prove, is not so easy for contemporary mathematicians using
> > |contemporary mathematics!
> > |
> > |Following from my research on factoring polynomials into
> > |non-polynomial factors I have a rather simple result that given prime
> > |integers, f_1, f_2, and integer M, where
> > |
> > |M = f_1 f_2
> > |
> > |if you find integers x and y, where
> > |
> > |x^j = -y^j mod M
> > |
> > |where j is a positive odd integer, and abs(x)>1 and abs(y)>1,
> > |
> > |then it must be true that
> > |
> > |(x+y) = 0 mod f_1 or
> > |
> > |(x+y) = 0 mod f_2.
> > |
> > |I'd like to see someone prove that result, if you can.
> > |
> > |I consider this a rather easy test.
> > |
> > |And to answer another poster who replied in my previous thread, yes, I
> > |have the proof.
> > |
> > |The question here is, do any of you?
> >
> > f1=7,f2=13,j=3,x=53,y=75
> >
>
> Well I'm impressed you found that counterexample in 76 minutes! Could
> you explain how you did it to those of us who don't know where to start?

You write a C program. This is just being typed in, without testing.
There may be a bug or two, and it only tries the case j=3.

#include <stdio.h>

// This function takes a number M, returns true if M is the product
// of exactly two primes which will be stored in *f1 and *f2; returns
// false if M is not the product of exactly two primes.
int two_factors (int M, int* f1, int* f2) {
int i, j, k;
for (i = 2; i*i <= M; ++i) {
if (M % i != 0) continue;
k = M / i;
for (j = i;j*j <=k; ++j) {
if (k % j == 0) break;
}
if (j*j > k) { *f1 = i; *f2 = j; return 1;
}
return 0;
}

int main (void) {
int M;
for (M = 4; M <= 10000; ++M) {
int f1, f2;
if (! two_factors (M, &f1, &f2)) continue;
int x, y;
for (x = 2; x <= M-2; ++x) {
for (y = 2; y <= M-2; ++y) {
if ((x*x*x + y*y*y) % M == 0) {
if (x+y % f1 != 0 && x+y % f2 != 0) {
printf ("f1=%d, f2=%d, j=3, x=%d, y=%d\n",
f1, f2, x, y);
}
}
}
}
}
return 0;
}