It is impossible to write the right implementation of the Jacobi JCA (Q / P) symbol calculation algorithm. First, the algorithm itself:

Login: q , p – integers.

Exit: Jackobi symbol value.

1) s = 0, u = q, v = p.

2) Calculate R – the smallest positive residue during the division u on v . Calculate the whole k & gt; = 0 and odd t , such as r = t * 2 ^ k . Calculate `s = s + k * (V ^ 2 – 1) / 8 + (t – 1) * (V – 1) / 4 (mod 2)

3) If T = 1 , then the Jacobi symbol is (- 1) ^ s . End.

4) if t & gt; = 3 , then u = V , v = t , go to step 2.

Sales:

# include & lt; iostream & gt;
int jacobi (int q, int p)
{
  int s = 0, u = q, v = p;
  int r, k, t;
  do {
    // Calculate R - the smallest positive residue when dividing U on V
    r = u% v;
    // Calculate the integer k & gt; = 0 and odd T: R = T * 2 ^ K
    k = t = 0;
    While (R% 2 == 0)
    {
      k ++; // Indicator Degree deception among R
      R & GT; & GT; = 1; // Delim R to 2
    }
    t = r; // in T is the result of dividing R to 2 ^ k
    S = (S + K * (V * V - 1) / 8 + (T - 1) * (V - 1) / 4)% 2;
    if (t == 1)
      Return (s)? eleven;
    // New iteration
    U = V;
    V = T;
  } While (T & GT; = 3);
}
INT MAIN () {
  STD :: COUT & LT; & LT; Jacobi (-104, 997) & lt; & lt; STD :: ENDL;
  Return 0;
}

here j (-104, 997) should be equal to -1, and my implementation issues 1.

Tests from Help Maple: Jacobi (12, 3) = 0 , jacobi (28, 21) = 0 , Jacobi (6, 11) = -1 , Jacobi (226, 135) = 1 , jacobi (26, 35) = -1 , Jacobi (-286, 4272943 ) = 1 , Jacobi (888, 1999) = -1 .

Supplement: In ternary terms, I was wrong. If s == 1 (true), then you need to return -1, because the indicator of the degree is odd.

Maybe mathematics will complement something?

Answer 1, Authority 100%

And let’s Book open.

Jacobi symbol properties

for odd p & gt; 1 :

1. j (a, p) = j (a% p, p)
2. j (1, p) = 1
3. j (-1, p) = (-1) ^{(P-1) / 2}
4. j (ab..l, p) = j (a, p) j (b, p) … j (l, p) , reference: j (ab ^{2 , p) = j (a, p)}
5. j (2, p) = (-1) ^{(p 2 – 1) / 8}

for Positive odd mutually simple p and q :

6. j (q, p) = (-1) ^{(P – 1) (Q-1) / 4 J (P, Q)} .

Neither in the algorithm, nor in the program, I do not even see hints of any analysis of signs or factorization.
Those. The main used formula looks unproved .

Also known that the Jacobi symbol J (A, P) is +1 or -1 , because is defined For all a , not dividing on p . Therefore, zero The result in the test Maple needs a special interpretation – for example, for J (12.3) .