Homepage Solution manuals Ivan Niven An Introduction to the Theory of Numbers Exercise 2.1.40 (Sum of the elements of a complete (resp. reduced) residue system)

Exercise 2.1.40 (Sum of the elements of a complete (resp. reduced) residue system)

For m odd, prove that the sum of the elements of any complete system modulo m is congruent to zero modulo m ; prove the analogous result for any reduced system for m > 2 .

Answers

Proof. Let { r 1 , r 2 , , r m } be a complete system modulo m , where m is odd, and S = r 1 + r 2 + + r m .

a)

First proof. Since a and m are relatively prime, Theorem 2.6 shows that 2 r 1 , 2 r 2 , , 2 r m is a complete system modulo m .

Therefore there is a permutation σ S n such that

2 r i = r σ ( i ) , i = 1 , , m .

Then

2 S = 2 r 1 + + 2 r m r σ ( 1 ) + + r σ ( m ) ( mod m ) ,

and r σ ( 1 ) + + r σ ( m ) = S .

Thus 2 S S ( mod m ) , so S 0 ( mod m ) .

Second proof. Since { r 1 , r 2 , , r m } be a complete system modulo m ,

pℤ = { [ r 1 ] , , [ r m ] } .

where [ a ] denotes the class of a in mℤ . Thus

[ S ] = x mℤ x = i = 0 m 1 [ i ] = [ m ( m 1 2 ) ] ( m  is odd. ) = [ m ] [ m 1 2 ] = [ 0 ] .

Therefore S 0 ( mod m ) .

b) Write ( mℤ ) × the invertible elements of mℤ , i.e. the elements [ i ] mℤ such that i m = 1 .

Let { r 1 , , r ϕ ( m ) } be a reduced system modulo m , where m > 2 , and S = r 1 + + r ϕ ( m ) , so that

( mℤ ) × = { [ r 1 ] , , [ r ϕ ( m ) ] } .

The proof of S 0 ( mod m ) is very different ( aS S ( mod m ) doesn’t imply S 0 ( mod m ) ).

Here m > 2 , so m m = m 1 , and m 2 ( mℤ ) × (if m is odd, m 2 , and if m is even ( m 2 ) m = m 2 1 ).

S 0 < i < m , i m = 1 i ( mod m ) .

Note that

0 < i < m 2 , i m = 1 m 2 < m i < m , ( m i ) m = 1 ,

so that we can associate to every element i such that 0 < i < m 2 , i m = 1 the element m i in the sum S . Since i + ( m i ) = m 0 ( mod m ) , we obtain S 0 ( mod m ) .

More explicitly,

S 0 < i < m , i m = 1 i 0 < i < m 2 , i m = 1 i + m 2 < i < m , i m = 1 i 0 < i < m 2 , i m = 1 i + 0 < j < m 2 , j m = 1 ( m j ) ( j = m i ) 0 < i < m 2 , i m = 1 i + 0 < i < m 2 , i m = 1 ( m i ) 0 < i < m 2 , i m = 1 i + ( m i ) 0 < i < m 2 , i m = 1 m 0 ( mod m ) .

So, for any reduced system { r 1 , , r ϕ ( m ) } , where m > 2 ,

r 1 + + r ϕ ( m ) 0 ( mod m ) .

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2024-08-02 08:02
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