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By Leo Moser

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M − 1, is obtained from a particular solution p1 (x) as follows. p(x) = p1 (x) + (x − a1 )(x − a2 ) · · · (x − ak−1 ) · Q(x), where Q(x) is any polynomial with integer coefficients. (3) 48 Chapter 5. Congruences Theorem 2. {pn (c)}, n ≥ 0, is relatively prime for all c if and only if p(x) belongs to one of the following six classes of polynomials. 1 + x(x − 1) · Q(x) 1 − x − x2 + x(x2 − 1) · Q(x) 1 − 2x2 + x(x2 − 1) · Q(x)) 2x2 − 1 + x(x2 − 1) · Q(x) x2 − x − 1 + x(x2 − 1) · Q(x) − 1 + x(x + 1) · Q(x) Proof.

Indeed, T means adding 1 to the complex number corresponding to the point and R means multiplication by eiθ . Hence all our points are polynomials in eiθ with positive coefficients, say z = P eiθ . But now if a point has a double representation, then P eiθ = R(eiθ ) and we would obtain a polynomial in eiθ which would negate the transcendental character of eiθ . Let T˜ denote the subset of S˜ which consists of those points of S˜ for which ˜ denote the subset the last operation needed to reach them is a T , and let R ˜ which consist of those points of S for which the last operation needed to reach ˜ and T˜ ∩ R ˜ = ∅.

The main thing to check here is that every element has an inverse in the system. If we seek an inverse for a we form aa1 , aa2 , . . , aaϕ(m) . We have already seen that these are ϕ(m) numbers incongruent mod m and relatively prime to m. Thus one of them must be the unit and the result follows. We now regain Euler’s proof from that of Lagrange’s which states that if a is an element of a group G of order m, then am = 1. In our case this means aϕ(m) ≡ 1 or ap−1 ≡ 1 if p is a prime. The integers under p form a field with respect to + and ×.

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