Dr. Marques Sophie
Office 519
Number theory
Fall Semester 2013
[email protected]
Problem Set #5
Due monday october 21th in Class
Exercise 1: (?) 4 points
Prove that there are infinitely many integers which can not be written as a sum of three
squares. (Hint: Consider mod 8.)
Solution:
First, we have the following table:
02
22
42
62
≡0
≡4
≡0
≡4
mod
mod
mod
mod
12
32
52
72
8
8
8
8
≡1
≡1
≡1
≡1
mod
mod
mod
mod
8
8
8
8
Thus, the possible values of squares mod 8 are 0, 1, and 4. Hence, the possible values
of sums of three squares mod 8 are:
0 + 0 + 0 ≡ 0 mod 8
0 + 0 + 4 ≡ 4 mod 8
0 + 1 + 4 ≡ 5 mod 8
1 + 1 + 1 ≡ 3 mod 8
1 + 4 + 4 ≡ 1 mod 8
0 + 0 + 1 ≡ 1 mod 8
0 + 1 + 1 ≡ 2 mod 8
0 + 4 + 4 ≡ 0 mod 8
1 + 1 + 4 ≡ 6 mod 8
4 + 4 + 4 ≡ 4 mod 8
From this table, we can conclude that any integer of the form 8k + 7 cannot be written
as a sum of three squares.
Exercise 2: (?) 4 points
Consider the rational number
Sn =
n
X
1
k=1
k
=
un
vn
where un and vn are positive integers. Prove that if p is an odd prime, then the numerator up−1 of Sp−1 is divisible by p. (Hint: Write Sp−1 as a fraction with denominator
(p − 1)!, and apply Wilson’s theorem.)
Solution:
First, note that a map (Z/p)× 3 k 7→ k −1 ∈ (Z/p)× is bijective. Thus, it follows from
1
Wilson’s theorem that
(p − 1)!Sp−1 =
p−1
X
(p − 1)!
k=1
p−1
≡−
X
k
k
−1
≡ (p − 1)!
p−1
X
k −1
mod p
k=1
≡−
k=1
p−1
X
k≡0
mod p
k=1
Exercise 3: (?) 4 points
1. Prove that n7 − n is divisible by 42, for any integer n. (Hint: Use Fermat’s little
theorem.)
2. n5 − 5n3 + 4n is divisible by 120, for every integer n (Hint: Factorize n5 − 5n3 + 4n
and 120.).
Solution:
1. By Fermat’s little theorem, n2 ≡ n mod 2. Hence n7 ≡ n mod 2. Also by Fermat’s
little theorem, n3 ≡ n mod 3. Hence
n7 ≡ n3 n3 n ≡ n3 ≡ n mod 3
Finally, by Fermat’s little theorem, one more time n7 ≡ n mod 7.
2. Let N = n5 − 5n3 + 4n. Then
n5 − 5n3 + 4n = n(n4 − 5n + 4) = n(n2 − 1)(n2 − 4) = n(n − 1)(n + 1)(n − 2)(n + 2)
So N is the product of the five consecutive integers: (n − 2), (n − 1), n, (n + 1),
(n + 2). Exactly one of these integers is divisible by 5, there are at least 2 distinct
even number with at least one is divisible by 4 and at least one is divisible by 3.
Hence, 120 = 8 × 5 × 3 divides N .
Exercise 4: (?) 4 points
If p is an odd prime, prove that
12 × 32 × 52 × ... × (p − 2)2 ≡ (−1)(p+1)/2 mod p
and
22 × 42 × 62 × ... × (p − 1)2 ≡ (−1)(p+1)/2 mod p
Solution:
22 ×42 ×62 ×...×(p−1)2 ≡ 1×(p−1)×3×(p−2)×...×(p−2)(p−(p−2))(−1)(p−1)/2 mod p
2
Since there are (p − 1)/2 numbers in this list 1, 3, ..., p − 2. Now,
1(p − 1)2(p − 2)...(p − 2)(p − (p − 2))(−1)p(p−1)/2 ≡ (p − 1)!(−1)(p−1)/2 mod p
Using Wilson’s theorem,
(p − 1)!(−1)(p−1)/2 ≡ (−1)(−1)(p−1)/2 ≡ (−1)(p+1)/2 mod p
The second part of the problem is done similarly.
Exercise 5: (??) 4 points
1. Prove that if n is an integer and p is prime then p|n or p|np−1 − 1.
2. If m, n are integer and p is a prime and l is a natural number then p divides
mn(ml(p−1) − nl(p−1) ).
(Hint: use Fermat’s little theorem.)
3. Show that mn(m60 − n60 ) is divisible by 56786730 for any m and n. (Hint:
566786730 = 2 × 3 × 5 × 7 × 11 × 13 × 31 × 61.)
Solution:
1. Observe np −n = n(np−1 −1). Hence the result follows immediately from Fermat’s
little theorem and Euclid’s lemma.
2. If p|m or p|n then the result is true. So, Suppose that p - m and p - n. Then by
the previous question p|mp−1 − 1 and p|np−1 − 1. Hence,
mp−1 ≡ 1 ≡ np−1 mod p
and so, for any natural number l,
ml(p−1) ≡ 1 ≡ nl(p−1) mod p
so that ml(p−1) − nl(p−1) ≡ 0 mod p, i.e.
p|ml(p−1) − nl(p−1)
So in all cases, p|mn(ml(p−1) − nl(p−1) ).
Now observe the following:
mn(m60 − n60 ) =
=
=
=
=
=
=
=
mn(m60(2−1) − n60(2−1) )
mn(m30(3−1) − n30(3−1) )
mn(m15(5−1) − n15(5−1) )
mn(m12(7−1) − n12(7−1) )
mn(m10(11−1) − n10(11−1) )
mn(m5(13−1) − n5(13−1) )
mn(m2(31−1) − n2(31−1) )
mn(m1(61−1) − n1(61−1) )
3
1
1
(?) = easy , (??)= medium, (???)= challenge
4