111 Solutions of Polynomial Congruences Let f be a polynomial with integer

111 solutions of polynomial congruences let f be a

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1.11 Solutions of Polynomial Congruences Let f be a polynomial with integer coefficients, and let m be a positive integer. If x and x 0 are integers with x x 0 (mod m ) then f ( x ) f ( x 0 ) (mod m ). It follows that the set of integers x satisfying the congruence f ( x ) 0 (mod m ) is a union of congruence classes modulo m . The number of solutions modulo m of the congruence f ( x ) 0 (mod m ) is defined to be the number of congruence classes of integers modulo m such that an integer x satisfies the congruence f ( x ) 0 (mod m ) if and only if it belongs to one of those congruence classes. Thus a congruence f ( x ) 0 (mod m ) has n solutions modulo m if and only if there exist n integers a 1 , a 2 , . . . , a n satisfying the congruence such that every solution of the congruence f ( x ) 0 (mod m ) is congruent modulo m to exactly one of the integers a 1 , a 2 , . . . , a n . Note that the number of solutions of the congruence f ( x ) 0 (mod m ) is equal to the number of integers x satisfying 0 x < m for which f ( x ) 0 (mod m ). This follows immediately from the fact that each congruence class of integers modulo m contains exactly one integer x satisfying 0 x < m . Theorem 1.24 Let f be a polynomial with integer coefficients, and let p be a prime number. Suppose that the coefficients of f are not all divisible by p . Then the number of solutions modulo p of the congruence f ( x ) 0 (mod p ) is at most the degree of the polynomial f . 13
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Proof The result is clearly true when f is a constant polynomial. We can prove the result for non-constant polynomials by induction on the degree of the polynomial. First we observe that, given any integer a , there exists a polynomial g with integer coefficients such that f ( x ) = f ( a ) + ( x - a ) g ( x ). Indeed f ( y + a ) is a polynomial in y with integer coefficients, and therefore f ( y + a ) = f ( a )+ yh ( y ) for some polynomial h with integer coefficients. Thus if g ( x ) = h ( x - a ) then g is a polynomial with integer coefficients and f ( x ) = f ( a ) + ( x - a ) g ( x ). Suppose that f ( a ) 0 (mod p ) and f ( b ) 0 (mod p ). Let f ( x ) = f ( a ) + ( x - a ) g ( x ), where g is a polynomial with integer coefficients. The coefficients of f are not all divisible by p , but f ( a ) is divisible by p , and therefore the coefficients of g cannot all be divisible by p . Now f ( a ) and f ( b ) are both divisible by the prime number p , and therefore ( b - a ) g ( b ) is divisible by p . But a prime number divides a product of integers if and only if it divides one of the factors. Therefore either b - a is divisible by p or else g ( b ) is divisible by p . Thus either b a (mod p ) or else g ( b ) 0 (mod p ). The required result now follows easily by induction on the degree of the polynomial f . 1.12 Primitive Roots Lemma 1.25 Let m be a positive integer, and let x be an integer coprime to m . Then there exists a positive integer n such that x n 1 (mod m ). Proof There are only finitely many congruence classes modulo m . Therefore there exist positive integers j and k with j < k such that x j x k (mod m ).
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