试证

用a[i,j] (1 ≤ i,j ≤ 2n+1) 表示格中第i行j列的值，则如下赋值是一个解

a[i,j] = [(n + i - j) % (2n+1)] * (2n+1) + [(i+j-(n+2)) % (2n+1)] + 1

  %表示取余运算  例如 7%5=2 -2%5=3  5%5=0 18%5=3

证明： 我们需要证明如下3点：
1)  1 ≤ a[i,j] ≤ (2n+1)^2
2)  a[i,j]两两不同
3)  Sigma(j=1 to 2n+1) a[i,j]             //每行
   = Sigma(i=1 to 2n+1) a[i,j]            //每列
   = Sigma(i=1 to 2n+1) a[i,i]            //对角线1
   = Sigma(i=1 to 2n+1) a[i,(2n+1)-i+1]   //对角线2
   = 某一定值

1) 1 ≤ [(n + i - j) % (2n+1)] * (2n+1) + [(i+j-(n+2)) % (2n+1)] + 1
     <= (2n+1-1)*(2n+1) + (2n+1-1) + 1
     = 2n(2n+1) + 2n + 1
     = (2n+1)^2  

2) 假若有1到2n+1之间的4个整数i1,j1, i2,j2满足
   a[i1,j1] = a[i2,j2]   
  则
   [(n + i1 - j1) % (2n+1)] * (2n+1) + [(i1 + j1-(n+2)) % (2n+1)] + 1
   = [(n + i2 - j2) % (2n+1)] * (2n+1) + [(i2 + j2-(n+2)) % (2n+1)] + 1
  因此
     [(i1-i2) - (j1-j2)) % (2n+1)] * (2n+1) + [((i1-i2) + (j1-j2)) % (2n+1)] = 0
 令  Δi=i1-i2,  Δj=j1-j2,
 则  [( Δi - Δj) % (2n+1)] * (2n+1) + [(Δi + Δj) % (2n+1)] = 0
  因 (Δi - Δj) % (2n+1) ≥ 0,  (Δi + Δj) % (2n+1) ≥ 0, 必有
   (Δi - Δj) % (2n+1) = 0
   (Δi + Δj) % (2n+1) = 0
  因此 Δi - Δj = u*(2n+1)
       Δi + Δj = v*(2n+1)

u,v为整数
Δi = (u+v)*(2n+1)/2

Δj = (v-u)*(2n+1)/2
 因2n+1是奇数，(u+v)和(u-v)必为偶数，即
   Δi = s*(2n+1)
   Δj = t*(2n+1)
  s，t为整数
 又因 1 ≤ i1, i2 ≤ 2n+1
  -2n = (1-(2n+1) ≤  Δi = i1 - i2 ≤ (2n+1-1) = 2n
  必有 Δi=0
  同理 Δj=0
  即i1=i2, j1=j2
 
3) 令N = Sigma(i=1 to (2n+1)^2)/(2n+1) = (2n+1) * (2n^2 + 2n + 1), 注意N是只依赖于n的定值
 Sigma(j=1 to 2n+1) a[i,j]
  = Sigma(j=1 to 2n+1) {[(n + i - j) % (2n+1)] * (2n+1) + [(i+j-(n+2)) % (2n+1)] + 1}
  = (2n+1) * Sigma(j=1 to 2n+1) [(n + i - j) % (2n+1)] + Sigma(j=1 to 2n+1) [(i+j-(n+2)) % (2n+1)] + (2n+1)
  = (2n+1) * (0+1+...+2n) + (0+1+...+2n) + (2n+1)
  = (2n+1) * 2n * (2n+1)/2 + 2n * (2n+1)/2 + (2n+1)
  = (2n+1) * (2n^2+n + n + 1)
  = (2n+1) * (2n^2 + 2n + 1)
  = N
 
  同样可得
   Sigma(i=1 to 2n+1) a[i,j]
   =(2n+1) * (2n^2 + 2n + 1)
   = N
   
  Sigma(i=1 to 2n+1) a[i,i]
  = Sigma(i=1 to 2n+1) {[(n + i - i) % (2n+1)] * (2n+1) + [(i+i-(n+2)) % (2n+1)] + 1}
  = Sigma (i=1 to 2n+1) [n*(2n+1)] + Sigma (i=1 to 2n+1) [(2i-(n+2)) % (2n+1)] + (2n+1)
  = (2n+1)^2 * n + (0+1+...+2n) + (2n+1)
     // 注意： 当i从1到2n+1遍历时， 2i(从而2i-(n+2))除以2n+1的余数遍历0到2n
  = (2n+1)^2 * n + 2n*(2n+1)/2 + (2n+1)
  = (2n+1)(2n^2 + 2n + 1)
  = N
 
  Sigma(i=1 to 2n+1) a[i,(2n+1)-i+1]
 = Sigma(i=1 to 2n+1) {[(n + i - (2n+1)+i-1) % (2n+1)] * (2n+1) + [(i+(2n+1)-i+1-(n+2)) % (2n+1)] + 1}
 = Sigma(i=1 to 2n+1) [(2i - (n+1)-1) % (2n+1)] * (2n+1) + Sigma(i=1 to 2n+1)[((2n+1)+1-(n+2)) % (2n+1)] + (2n+1)
 = (0+1+...+2n)*(2n+1) + (2n+1)*n + (2n+1)
 = (2n+1)^2*2n/2 + (2n+1)*n + 1
 = (2n+1)*(2n^2+n + n + 1)
 = (2n+1)*(2n^2+2n+1)
 = N