The Theory of Numbers by Robert Daniel Carmichael
By Robert Daniel Carmichael
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A textual content in accordance with classes taught effectively over a long time at Michigan, Imperial university and Pennsylvania nation.
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Additional info for The Theory of Numbers
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.