| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
The following problem of mine was rejected by the Mathematics
Magazine as being too easy.
Let Z be the integers modulo p and let C denote the complex numbers.
p
2 2 2 2
The quadratic polynomial x + y + z factors over Z [x,y,z] as (x+y+z) .
2
2 2 2
(a) Does x + y + z factor as the product of two linear polynomials
over Z [x,y,z] for any prime p>2?
p
2 2 2
(b) Does x + y + z factor as the product of two linear polynomials
over C[x,y,z] ?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 72.1 | It's about time | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Tue Apr 23 1991 15:19 | 19 |
Goodness, this is an oldie for it not to have been answered. > 2 2 2 >(a) Does x + y + z factor as the product of two linear polynomials > > over Z [x,y,z] for any prime p>2? > p No. Let x^2+y^2+x^2 = (ax+by+cz)*(dx+ey+fz). Then we get ad = 1, be = 1, cf = 1 (mod p) and ae+bd=0, af+cd=0, bf+ce=0 (mod p). Eliminating d,e,f from these leaves a^2+b^2=0, a^2+c^2=0, c^2+b^2=0 (mod p) or 2a^2=2b^2=2c^2=0 mod p, so p is even, therefore p is 2. I think the same argument holds for complex coeff's as well. | |||||