We found \ (p (p-1) \) within a range, each number about \ (P \) and \ (\ varphi (p) \ ) of the remainder of the \ ((a, b) \ ) varies .
Set original roots \ (G \) .
\ [n-\ equiv G ^ {A} (MOD \ P) \]
\ [m \ equiv G ^ {B} (MOD \ P) \]
\ [n-\ equiv C (MOD \ \ varphi (P)) \ ]
\ [m \ equiv D (MOD \ \ varphi (P)) \]
so in other words, the \ (n, m \) is not equal to 0:
\ [AD \ equiv BC (MOD \ \ varphi ( p)) \]
Because of the remainder varied.
So in fact is seeking:
equiv CD, A, B, C, D \ in [0, \ varphi (P)) \] ((P) MOD \ \ varphi) \ [ab & \
Number of Solutions.
If \ (n, m \) is equal to 0, equal to the solution portion 0 is \ ((p-1) ^
2 \) is now provided:
\ (N (m) \) of \ (m \) satisfies the conditions the number of four arrays.
Then set up:
\ [m = * \ _ p
^ e \] Since:
set \ (p_i ^ {e_i} \ ) one solution and \ (p_j ^ {e_j} \ ) one solution are, respectively:
\ [a_ib_i \ equiv c_id_i (MOD \ P_i ^ {e_i}) \]
\ [a_jb_j \ equiv c_jd_j (MOD \ p_j ^ {e_j}) \]
so obtained:
\ [a_ia_jb_ib_j \ equiv c_ic_jd_id_j (MOD \ P_i ^ {e_i} p_j ^ {e_j }) \]
this way we obtained two prime numbers:
\ [N (e_i P_i} ^ {^ {e_j p_j}) = N (P_i} ^ {e_i) N (p_j e_j ^ {}) \]
so:
\ [N (m) = \ Prod \ limits_ {I =. 1} ^ {W} N (P_i ^ {e_i}) \]
we now require: \ (N (P ^ E) \)
we set:
\ [ab \ equiv t (mod
\ p ^ e), a, b \ in [0, p ^ e) \] a number satisfying the \ ((a, b) \ ) number is: \ (C (T ) \)
then:
\ [N (P ^ E) = \ SUM \ limits_ {I = 0} ^ {P ^ E-. 1} C ^ 2 (I) \]
now consider how to find \ (C (t) \ )
suppose we have now identified\ (A \) , required to satisfy the conditions \ (B \) number.
But we find that \ (gcd (a, p ^ e) | t \) This condition is not necessarily true, that is to say the number is likely to meet the conditions \ (0 \) , so we have convenient batch computing, is still first assumed \ (ab! = 0 \) after case, 0 special sentence out.
Therefore proposed here \ (A \) contained in the (P \) \ factor.
Conditions are converted to:
\ [^ JS T = P, A = P ^ {\ Alpha} A ', B = P ^ {\ Beta} B', A'B '\ equiv S (MOD \ P ^ {} EJ ) \]
and we need to solve is to meet certain \ (\ alpha, \ beta \ ) in the case where the \ ((a ', b' ) \) number.
Assume \ (a '\) have been identified.
Then:
\ [A'B '\ equiv S (MOD \ P ^ EJ {}) \]
is converted to a linear equation:
\ [A'B' + KP ^ {S =} EJ, B \ in [0, P ^ {e- \ beta}) \]
first Lemma:
provided a'b 'to a group of specialized solution is: \ [\} b_0 the begin {\\ k_0 Cases \ Cases End {} \]
Set \ (d = gcd (a '
, e ^ {ej}) \) then the general solution can be expressed as:
\ [\ the begin {Cases} B = b_0 \ FRAC {S} {D} + R & lt \ FRAC {P ^ { ej}} {d} \\ k
= k_0 \ frac {s} {d} -r \ frac {a '} {d} \ end {cases} \] where \ (R & lt \) runs over integers.
Prove:
we set the equation \ (a '(b' + g_b) + p ^ {ej} (k-g_k) = s \) was established.
Then:
\ [A'B '+ P + a'g_b EJ ^ {} {^ EJ} G_k KP = S \]
\ [^ {P = EJ a'g_b} G_k \]
\ [{A'} = _ ddg_b EJ P ^ {} {} _ ddg_k \]
\ [\ P ^ {EJ FRAC {}} {D} | G_B, \ FRAC {a '} {D} | G_k \]
which can verify the above lemma complete.
With this enough to say.
This equation:
\ [GCD (A ', E ^ P) =. 1 \]
so \ (b' = b_0 + rp
^ {ej} \) That is, each \ (p ^ {ej} \ ) appears a solution.
And interval values:
\ [[0, P ^ {\ Beta}) \]
So for the same \ (A '\) ,\ (b '\) Number is: \ (\ FRAC {P ^ {\ {Beta}} ^ {P E- \ beta-\ Alpha}} = {P ^ \ Alpha} \) .
Our \ (a '\) No \ (P \) factor, his number is: \ (\ E- FRAC {P ^ {\ Alpha} (. 1-P)} = {P ^ {P} e- \ alpha-1} (p
-1) \) so that for a certain group \ ((\ Alpha, \ Beta) \) , \ (a ', B' \) number is:
\ [^ {P E - \ alpha-1} (p
-1) p ^ {\ alpha} = p ^ {e-1} (p-1) \] while \ (j + 1 \) group \ ((\ alpha, \ beta ) \)
Therefore: for one \ (T = P ^ JS \) : \
[C (T) = (J +. 1) P ^ {E-. 1} (P-. 1) \]
now Laid determination 0:
\ (1.a = 0 \!)
then: \ (ab & E ^ + KP = 0 \)
disposed \ (a = p ^ {\
alpha} g \) then \ (d = gcd (a, p ^ e) = p ^ {\ alpha} \)
\ (b = b_0 \ frac {
0} {d} + r \ frac {p ^ {e}} {d} = r \ frac {p ^ e} {d} \) and: \ (B \ in [0 , p ^ e) \)
so that the number of said solution is: \ (\ E FRAC {P ^ {} \ FRAC P ^ {E}}} = {D = D {P ^ \ Alpha} \)
satisfies the \ (gcd (a, p ^ e ) = p ^ {\ alpha} \) a \ (a \) number is: \ (\ E FRAC {P ^ {P ^ {} \ Alpha}} \ {P FRAC -1} {p} \)
to \ (C (0) \) the contribution is: \ [^ {P \ Alpha} \ FRAC {E} P ^ {P ^ {\ Alpha}} \ {P-FRAC. 1 } {p} = p ^ {
e-1} (p-1) \] and \ (\ alpha \ in [0
, e-1] \) then the contribution of this part is: \ (EP ^ {E-. 1 } (. 1-P) \)
\ (2.a == 0 \)
contribution to this part is \ (B \) number as \ (p ^ e \)
Then:
\ (C (0) = E-EP. 1} ^ {(. 1-P) + E ^ = P ^ {EP}. 1-E (. 1-P) + P-E. 1} ^ {((p- 1) +1) = ep ^ { e-1} (p-1) + p ^ {e-1} (p-1) + p ^ {e-1} = (e + 1) p ^ {e- 1} (p-1) +
p ^ {e-1} \) Therefore, considering the \ (J \) identical contributions are the same:
\ [N (P ^ E) = C ^ 2 (0) + \ SUM \ limits_ {i = 0} ^
{e-1} p ^ {ej-1} (p-1) C ^ 2 (j) \] this equation may be directly iteration \ (log \) the complexity of the solution.
It can be determined \ (N (m) \)
\ (ans = (p-1
) ^ 2 + N (varphi (p)) \) and for \ (varphi (p) \) decomposition can first screening out \ (\ sqrt {p} \ ) primes within.
Set \ (P \) number of prime numbers in parentheses are \ (\ pi (p) \
) then the complexity:
\ [O (\ sqrt {P} + T (\ PI (\ sqrt {P}))) \ ]
the problem is resolved.