论文标题
$ \ prod f（i，j）$ 0 <i <j <j <p/2，\ p \ nmid f（i，j）$超过$ \ prod f（i，j）$
Legendre Symbol of $\prod f(i,j) $ over $ 0<i<j<p/2, \ p\nmid f(i,j) $
论文作者
论文摘要
令$ p> 3 $为素数。我们调查了$ \ displayStyle \ prod_的legendre符号{0 <i <j <j <p/2 \ at p \ nmid f（i，j）} f（i，j）} f（i，j）\ $，其中$ i，j \ in \ in \ bbb z，f（i，j）$是带有integer系数的线性或二次形式。当$ f = ai^2+bij+cj^2 $和$ p \ nmid c（a+b+c）$时，我们证明，评估产品等同于确定$ \ displayStyle \ sum_ = sum_ {y = 1}^{p-1}^{p-1} \ bigG（ \ pmod {16} $，其中$ 4C（a+b+c）k \ equiv（4ac-b^2）\ pmod {p}。$ parallel结果以$ \ displayStyle \ prod_ {i，j = 1，j = 1 \ atop p \ atop p \ nmid f（I， $ and}。对于几类F（i，j），可以明确评估这两种产品。最后，当F是线性形式时，我们为这些产品提供了统一的身份。因此，我们证明了在太阳报中提出的这类问题。
Let $p>3$ be a prime. We investigate Legendre symbol of $\displaystyle \prod_{0<i<j<p/2 \atop p\nmid f(i,j) } f(i,j) \ $, where $i,j\in \Bbb Z, f(i,j)$ is a linear or quadratic form with integer coefficients. When $f=ai^2+bij+cj^2$ and $p\nmid c(a+b+c)$ , we prove that to evaluate the product is equivalent to determine $ \displaystyle \sum_{y=1}^{p-1} \bigg(\frac{y(y+1)(y+k)}{p}\bigg) \pmod{16}$ , where $4c(a+b+c)k \equiv (4ac-b^2)\pmod{p}.$ Parallel results are given for $\displaystyle \prod_{i,j=1 \atop p\nmid f(i,j) }^{(p-1)/2} \bigg(\frac{ f(i,j) }{p}\bigg).$ Then we show that $ \displaystyle \sum_{y=1}^{p-1} \bigg(\frac{y(y+1)(y+k)}{p}\bigg) \pmod{16}$ can be evaluated explicitly when k=2,4,5,9,10 or k is a square. And for several classes of f(i,j) these two kinds of products can be evaluated explicitly. Finally when f is a linear form we give unified identities for these products. Thus we prove these kind of problems raised in Sun's paper.
