자바 스크립트 (ES7), 165 바이트

@NeilA.의 답변 과 마찬가지로 이것은 Tunnell의 정리를 기반으로 하므로 Birch 및 Swinnerton-Dyer 추측이 사실이라고 가정합니다.

부울 값을 반환합니다.

n=>(r=(g=i=>i<n?g(i+!(n%i**2?0:n/=i*i)):n**.5|0)(s=2),g=(C,k=r)=>k+r&&g(C,k-1,C(k*k)))(x=>g(y=>g(z=>s+=2*(n==(X=(n&1?2:8)*x+(o=2-n%2)*y)+o*32*z)-(n==X+o*8*z))))|s==2

온라인으로 사용해보십시오!

방법?

n

$n$n′

$n'$r=⌊n′−−√⌋

$r=\lfloor\sqrt{n'}\rfloor$s

$s$2

$2$

r = (                // we will eventually save isqrt(n) into r
  g = i =>           // g = recursive function taking an integer i
    i < n ?          //   if i is less than n:
      g(i + !(       //     do a recursive call with either i or i + 1
        n % i**2 ?   //     if n is not divisible by i²:
          0          //       yield 0 and therefore increment i
        :            //     else:
          n /= i * i //       divide n by i² and leave i unchanged
      ))             //     end of recursive call
    :                //   else:
      n ** .5 | 0    //     stop recursion and return isqrt(n)
  )(s = 2)           // initial call to g with i = s = 2

g

$g$C

$C$k2

$k^2$−r<k≤r

$-r<k\le r$

  g = (C, k = r) =>  // C = callback function, k = counter initialized to r
    k + r &&         //   if k is not equal to -r:
    g(               //     do a recursive call:
      C,             //       pass the callback function unchanged
      k - 1,         //       decrement k
      C(k * k)       //       invoke the callback function with k²
    )                //     end of recursive call

g

$g$(x,y,z)∈[−r+1,r]3

$(x,y,z) \in [-r+1,r]^3$s

$s$2An=Bn

$2A_n = B_n$n

$n$2Cn=Dn

$2C_n=D_n$n

$n$

$$\begin{align} &A_n=\#\{(x,y,z)\in [-r+1,r]^3 \mid n=2x^2+y^2+32z^2\}\\ &B_n=\#\{(x,y,z)\in [-r+1,r]^3 \mid n=2x^2+y^2+8z^2\}\\ &C_n=\#\{(x,y,z)\in [-r+1,r]^3 \mid n=8x^2+2y^2+64z^2\}\\ &D_n=\#\{(x,y,z)\in [-r+1,r]^3 \mid n=8x^2+2y^2+16z^2\} \end{align}$$

g(x =>                            // for each x:      \    NB:
  g(y =>                          //   for each y:     >-- all these values are
    g(z =>                        //     for each z:  /    already squared by g
      s +=                        //       add to s:
        2 * (                     //         +2 if:
          n == (                  //           n is equal to either
            X =                   //           An if n is odd (o = 1)
            (n & 1 ? 2 : 8) * x + //           or Cn if n is even (o = 2)
            (o = 2 - n % 2) * y   //
          ) + o * 32 * z          //
        ) - (                     //         -1 if:
          n == X + o * 8 * z      //           n is equal to either
        )                         //           Bn if n is odd
    )                             //           or Dn if n is even
  )                               //
)                                 // if s in unchanged, then n is (assumed to be) congruent

답변

->n{[8,32].product(*[(-n..-t=1).map{|i|i*=i;n%i<1&&n/=i;i}*2+[0]]*3).map{|j|d=2-n%2
k,x,y,z=j
2*d*x+y+k*z==n/d&&t+=k-16}
t==1}

->n{t=0
[8,32].product(q=(-n..-1).map{|i|i*=i;n%i<1&&n/=i;i}*2+[0],q,q).map{|j|d=2-n%2
k,x,y,z=j
2*d*x+y+k*z==n/d&&t+=k-16}
t==0}

2*d*x^2 + y^2 + k*z^2 == n/d  where d=2 for even n and d=1 for odd n

->n{t=0                              #counter for solutions

  q=(-n..-1).map{|i|i*=i;n%i<1&&n/=i #make n square free by dividing by -n^2 to -1^2 as necessary 
  i}*2+[0]                           #return an array of squares, duplicate for 1^2 to n^2, and add the case 0 

  [8,32].product(q,q,q).map{|j|      #make a cartesian product of all possible values for k,x,y,z and iterate
    d=2-n%2                          #d=1 for odd n, 2 for even n
    k,x,y,z=j                        #unpack j. k=8,32. x,y,z are the squared values from q.
    2*d*x+y+k*z==n/d&&t+=k-16}       #test if the current values of k,x,y,z are a valid solution. If so, adjust t by k-16 as explained above.
t==0}                                #return true if t is the same as its initial value. otherwise false.