Beautiful polynomial

Algebra Level 5

Let P ( x ) P(x) is a polynomial with integer coefficients such that: P ( 1 ) = P ( 3 ) = P ( 5 ) = P ( 7 ) = a ; P ( 2 ) = P ( 4 ) = P ( 6 ) = P ( 8 ) = a P(1)=P(3)=P(5)=P(7)=a; P(2)=P(4)=P(6)=P(8)=-a Find the minimum positive value of a a .


The answer is 315.

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1 solution

Since P ( 1 ) = P ( 3 ) = P ( 5 ) = P ( 7 ) = a P(1)=P(3)=P(5)=P(7)=a , there exist a polynomial Q ( x ) Q(x) with integer coefficients and P ( x ) = Q ( x ) ( x 1 ) ( x 3 ) ( x 5 ) ( x 7 ) + a P(x)=Q(x)(x-1)(x-3)(x-5)(x-7)+a Put x = 2 , 4 , 6 , 8 x=2, 4, 6, 8 in above equality, we get: 2 a = 15 Q ( 2 ) = 9 Q ( 4 ) = 15 Q ( 6 ) = 105 Q ( 8 ) -2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8) .

From this, 2 a 2a is divisible by 15 , 9 15,9 , and 105 105 , so a a is divisible by 315 315 .

Hence, the minimum positive value of a a is 315 315 .

The equality holds iff Q ( 2 ) = Q ( 6 ) = 42 , Q ( 4 ) = 70 , Q ( 8 ) = 6 Q(2)=Q(6)=42,Q(4)=-70,Q(8)=-6 .

We can choose Q ( x ) = 8 x 3 + 124 x 2 576 x + 762 Q(x)=-8x^3+124x^2-576x+762 and P ( x ) = ( 8 x 3 + 124 x 2 576 x + 762 ) ( x 1 ) ( x 3 ) ( x 5 ) ( x 7 ) + 315 P(x)=(-8x^3+124x^2-576x+762)(x-1)(x-3)(x-5)(x-7)+315 .

Spoke too quickly. I only get 1 factor of 3, not 2.


Somewhat more directly (but essentially equivalent), we can use the fact that x y P ( x ) P ( y ) x-y \mid P(x) - P(y) to conclude that 3 × 5 × 7 2 a 3 \times 5 \times 7 \mid 2a .

Calvin Lin Staff - 4 years ago

How to find Q(x) ?

Kushal Bose - 4 years ago

this is an old AMC 12 problem actually. From 2010B I think.

Razzi Masroor - 1 year, 8 months ago

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