By Arun-Kumar S.

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Proof: 1. ) 2. ) 3. f (u) ≡pαi f (ai ) ≡pαi 0 from the fact that ai is a solution f (u) ≡pαi 0. i i i i 4. it means that ∀i pα i |f (u). 5. k i=1 i pα i |f (u) implies that m|f (u) 6. m|f (u) implies that f (u) ≡m 0 ✷ With that proof our problem of finding a solution to f (x) ≡m 0 reduces to a problem of finding a solution to f (x) ≡pαi 0, where p is a prime. 3 if f (x) ≡pαi 0 has a solution u then u is a solution of f (x) ≡pβ 0 for all 1 ≤ β ≤ α. 4 f (x) = n i i=1 ai x , where an = 0 then the kth derivative of f is a polynomial with degree ≤ n − k.

Qk ⇒ p = qi for some 1 ≤ i ≤ k, where q1 , q2 , . . , qk are all primes. 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 +.

Therefore, the exponent of p in r(p) 2n n is r(p) {[2n/pj ] − 2[n/pj ]} ≤ j=1 1 = r(p) j=1 The last inequality holds as each term in curly brackets is either 0 or 1. Taking the product over primes p ≤ 2n, we get the desired result. 4. If p satisfies 2n/3 < p ≤ n, then p occurs once in the prime factorization of n! and twice in (2n)! (as 2n 3p > 2n), hence as p > 2, p . n 5. This is proved by complete induction. Let P (n) denote the proposition to be proved. Clearly P (1), P (2) and P (3) hold, and if m > 1, we have P (2m) as: p < 42m−1 < 42m p= p≤2m p≤2m−1 42 CHAPTER 8.