Exercise 1.4.18* (Integer-valued polynomials, the end!)

Suppose that f ( x ) is an integer-valued polynomial of degree n and that g = g . c . d . ( f ( 0 ) , f ( 1 ) , , f ( n ) ) . Show that g f ( k ) for all integers k .

Answers

Proof. Write f ( x ) = j = 0 n c j ( x j ) . As in Exercise 15, ( c 0 , c 1 , , c n ) is solution of the linear system

f ( 0 ) = c 0 f ( 1 ) = c 0 + c 1 f ( 2 ) = c 0 + 2 c 1 + c 2 f ( i ) = c 0 + ( i 1 ) c 1 + + ( i i 1 ) c i 1 + c i f ( n ) = c 0 + ( n 1 ) c 1 + ( n 2 ) c 2 + ( n n 1 ) c n 1 + c n .

Let g = f ( 0 ) f ( 1 ) f ( n ) . Then g f ( i ) for all i [ [ 0 , n ] ] .

Note that g c 0 = f ( 0 ) . Reasoning by induction, assume that g c 0 , g c 1 , , g c i 1 for some i , 1 i n . Then

g f ( i ) [ c 0 + ( i 1 ) c 1 + + ( i i 1 ) c i 1 ] = c i .

The induction is done, so

i [ [ 0 , n ] ] , g c i .

Therefore, for any integer k , using ( k j ) (even if k = l < 0 : then ( k j ) = ( 1 ) j ( l 1 + j j ) ),

g f ( k ) = j = 0 n c j ( k j ) .

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2024-07-27 17:10
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