By Arun-Kumar S.

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I this is equivalent to the following claims. 1 if u is a solution of f(x) ≡m 0 then u is a solution of every equation f(x) ≡pαi 0. 2 if f(x) ≡pαi 0 has no solutions for some i, 1 ≤ i ≤k then f(x)≡m 0 has no solutions. 3 if each of f(x) ≡pαi 0 has solutions a1i , a2i , . . aki i which are all mutually incongurent solutions then i take u as any linear combination of solutions u ≡m resulting value u is a solution of f(x) ≡m 0. k i=1 Proof: proof for the first claim is if f(x) ≡m 0 has a solution u then 1.

Thus α satisfies a quadratic equation. Since the ai are all integers, the number [a0 , a1 , . ] = = [a0 , a1 , . . , an , α] 1 a0 + 1 a1 + a2 +···+α can be expressed as a polynomial in α with rational coefficients, so [a0 , a1 , . ] also satisfies a quadratic polynomial. Finally, α ∈ Q because periodic continued fractions have infinitely many terms. 36 CHAPTER 7. 3 The CF expansions of a qudratic irrationals x is purely periodic iff x<0 and −1 ≤ x < 0 P roof : (⇐=) Assume x > 1 xi+1 = 1 xi − ai ; 1 xi+1 x>1 and −1 ≤ = xi − ai as x = [a0 , .

We are used to considering primes only on natural numbers. Here is another set of primes over a different set. Consider the set of all even numbers Ze . The set Ze has the following properties: • for all a, b, c ∈ Ze , a + (b + c) = (a + b) + c - associativity. • for all a ∈ Ze , there is an element −a ∈ Ze , such that a + 0 = 0 + a = a, and 0 ∈ Ze - identity element. that this set forms an abelian group since it satisfies associativity, has an identity element (0), and for every even number x ∈ Ze , the negation −e is the unique inverse element under the operation +.