The Theory of Numbers by Robert Daniel Carmichael

By Robert Daniel Carmichael

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These are called quadratic congruences. The problem of the solution of the quadratic congruence (1) can be reduced to that of the solution of a simpler form of congruence as follows: Congruence (1) is evidently equivalent to the congruence 4α2 z 2 + 4αβz + 4αγ ≡ 0 mod 4αµ. (1) But this may be written in the form (2αz + β)2 ≡ β 2 − 4αγ mod 4αµ. Now if we put 2αz + β ≡ x mod 4αµ and β 2 − 4αγ = a, 4αµ = m, (2) CHAPTER 4. THE THEOREMS OF FERMAT AND WILSON 43 we have x2 ≡ a mod m. (3) We have thus reduced the problem of solving the general congruence (1) to that of solving the binomial congruence (3) and the linear congruence (2).

III. If a = b mod m, then ca = cb mod m, c being any integer whatever. The proof is obvious and need not be stated. IV. If a ≡ b mod m, α ≡ β mod m, then aα ≡ bβ mod m; that is, two congruences with the same modulus may be multiplied member by member. For, we have a = b+c1 m, α = β +c2 m. Multiplying these equations member by member we have aα = bβ + m(bc2 + βc1 + c1 c2 m). Hence aα ≡ bβ mod m. A repeated use of this theorem gives the following result: V. If a ≡ b mod m, then an ≡ bn mod m where n is any positive integer.

Now let η be any solution of (1). Then f (η) ≡ a0 (η − a)α (η − b)β . . (η − l)λ ≡ 0 mod p. Since p is prime it follows now that some one of the factors η − a, η − b, . . , η − l is divisible by p. Hence η coincides with one of the solutions a, b, c, . . , l. That is, (1) can have only the n solutions already found. This completes the proof of the theorem. EXERCISES 1. Construct a congruence of the form a0 xn + a1 xn−1 + . . + an ≡ 0 mod m, a0 ≡ 0 mod m, having more than n solutions and thus show that the limitation to a prime modulus in the theorem of this section is essential.