This is just combinatorics.

Find the number of quadratic polynomials, a x 2 + b x + c ax^2 + bx + c , which satisfy the following conditions:

(a) a , b , c a,b,c are distinct;
(b) a , b , c { 1 , 2 , 3 , , 1999 } a,b,c \in \{1,2,3,\ldots,1999 \} , and
(c) x + 1 x+1 divides a x 2 + b x + c ax^2 + bx + c .

1997001 1997002 1996001 1996002

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1 solution

Patrick Corn
Mar 5, 2020

Condition (c) is equivalent to a + c = b . a+c=b. For a given value of b , b, there are exactly b 1 b-1 pairs of positive integers ( a , c ) (a,c) summing to b , b, but if b b is even, we have to throw out a = c = b / 2 , a=c=b/2, so there are b 2 b-2 solutions for even b b and b 1 b-1 solutions for odd b . b. So the total number of solutions is 0 + 0 + 2 + 2 + 4 + 4 + + 1996 + 1996 + 1998 = 4 ( 1 + 2 + + 998 ) + 1998 = 998 999 2 + 1998 = 2 99 9 2 = 1996002 . 0+0+2+2+4+4+\cdots+1996+1996+1998 = 4(1+2+\cdots+998)+1998 = 998\cdot 999 \cdot 2 + 1998 = 2 \cdot 999^2 = \fbox{1996002}.

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