Sep 032013
因
\[(x^2+xy+y^2)(z^2+zw+w^2)=(xz-yw)^2+(xz-yw)[wx+y(z+w)]+[wx+y(z+w)]^2\]
因此, 形如 \(x^2+xy+y^2\) 的数相乘, 所得的积仍为同样的形式.
这恒等式是如何想出来的? 秘密在于行列式, 把 \(x^2+xy+y^2\) 看成行列式
\begin{vmatrix}
x& y\cr
-y & x+y
\end{vmatrix}
Let \(f(x_1,x_2,\dotsc,x_n)\) be a homogeneous polynomial. Let
\[S=\{f(a_1,a_2,\dotsc,a_n)\mid a_1,a_2,\dotsc,a_n \in\Bbb Z\}.\]
If \(S\) satisfies the following condition: for all \(m,n\in S\), we have \(mn\in S\). Can we determine all the homogeneous polynomials \(f\)?
For example, \(x^n(n\in\Bbb N),x^2+n y^2(n\in\Bbb Z), x^2+xy+y^2,x^3+y^3+z^3-3xyz\), and \(x^2+y^2+z^2+w^2\) are all appropriate examples.