0.1
Two number theory problems
Consider next two problems [1, Problems 30, 32]:
Problem 0.1.1 Let R(n) denote the number of ways to write a natural number n as the sum
n = a + b of two primes (for example, R(3) = 0, R(4) = 1, R(10) = 3, since 10 = 3 + 7 = 5 + 5 =
7 + 3). Show that if p < q are two consecutive prime numbers, then the sum 2R(q − 2) + 3R(q −
3) + 5R(q − 5) + ... + pR(q − p) is divisible by q.
Problem 0.1.2 Let x1 < x2 < ... < xn be an arithmetic progression of natural numbers, with x1
and x2 relatively prime. Let us assume that the product of the terms in the progression has only
m different prime divisors, where m < n. Prove that xn−m
≤ (n − 1)!.
1
Here we show how to prove them in slightly more general forms.
0.2
Proofs
Fact 0.2.1 Consider torsion-free Abelian group A and let S ⊆ A – be some finite subset of A.
Let R(a) denote the number of ways to write an element a ∈ A asPthe sum a = s1 + s2 of two
elements from S. Consider any element s ∈ S and element α := s0 ∈S s0 R(s − s0 ). Prove that
3α ∈< s >= {0, s, 2s, 3s, . . .}.
Proof:
Consider triples {s1 , s2 , s3 } ∈ (S × S × S)/S3 , where s1 + s2 + s3 = s. Name this set as T .
Note that we have three cases :
1) Elements s1 , s2 , s3 are pairwise different, so we have next relations s − s1 = s2 + s3 = s3 + s2 ,
s − s2 = s1 + s3 = s3 + s1 , s − s3 = s2 + s1 = s1 + s2 .
2) s1 = s2 6= s3 , so we have next relations s − s1 = s2 + s3 = s3 + s2 , s − s2 = s1 + s3 = s3 + s1 ,
s − s3 = s1 + s2 .
3) s1 = s2 = s3 , so we have next relations s − s1 = s2 + s3 , s − s2 = s1 P
+ s3 , s − s3 = s2 + s1 .
So let’s find contributions of number of these relations in general sum s0 ∈S s0 R(s − s0 ).
In case 1 we have +2s1 + 2s2 + 2s3 = +2s.
In case 2 we have +2s1 + s3 = +s.
In case 3 we have +s1 = +s/3.
So
P if we0 sum all0 these additions, we get that α :=
s0 ∈S s R(s − s ) = #{{s1 , s2 , s3 } ∈ T |s1 6= s2 6= s3 }2s + #{{s1 , s2 , s3 } ∈ T |s1 = s2 6= s3 }s +
#{{s1 , s2 , s3 } ∈ T |s1 = s2 = s3 }s/3. So easy to see that 3α ∈< s >. Remark 0.2.2 So problem 0.1.1 follows from fact 0.2.1 if we consider A = Z and set S – as set
of primes 0 < p ≤ q.
Fact 0.2.3 Consider sequence x1 , x2 , . . . , xn of nonzero natural numbers, such that
gcd(xi , xj )|gcd(i, j), ∀i 6= j. Let y1 ≤ y2 ≤ . . . ≤ xn – be permutation of set x1 , x2 , . . . , xn .
Let given that product x1 . . . xn has only m different prime divisors, where m < n. Prove that
y1 y2 . . . yn−m ≤ (n − 1)!.
Proof:
Denote set X := {x1 , x2 , . . . , xn }.
Let Pm : 2N → N be function depending on natural m, such that for every subset X ⊆ N,
Pm (X) = #{x ∈ X : m|x}.
] + 1.
Lemma 0.2.4 For any natural number m we have inequality Pm (X ) ≤ [ n−1
m
1
This lemma is easy exercise.
Q
Let νp : 2N → N be function, such that νp (X) = max{i : pi | x∈X x} for every P
set X ⊆ N and
prime
So from lemma 0.2.4 one can get that for every prime number p, νp (X ) = ∞
k=1 Ppk (X ) ≤
P∞ p.
n−1
[
]
+
max
ν
(x
)
=
ν
((n
−
1)!)
+
max
ν
(x
).
Product
x
.
.
.
x
has
only
m
different
prime
i p i
p
i p i
1
n
k=1 pk
Q
divisors, so p|x1 ...xn pmaxi νp (xi ) ≤ yn−m+1 yn−m+2 . . . yn .
Q
So from these properties of set X one can easily get that y1 y2 . . . yn = x1 x2 . . . xn = x∈X x ≤
(n − 1)!yn−m+1 yn−m+2 . . . yn . Remark 0.2.5 So problem 0.1.2 follows from fact 0.2.3 if we consider x1 < x2 < . . . < xn as
arithmetic progression with gcd(x1 , x2 ) = 1.
2
Bibliography
1. Olympiad, http://www.mathschool.ru/storage/SCContent/712/2014_en_problems_.pdf
3