The function f has the property that, for each real number x, f(x)+f(x-1)=x^2. If f(19)=94, what is the remainder when f(94) is divided by 1000?

1994 AIME Problem 3

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The function

f

has the property that, for each real number

x,

f(x)+f(x-1)=x^2.

f(19)=94,

what is the remainder when

f(94)

is divided by

1000?

Answer

First use

x=20

to find

f(20)

Eliminate alternating terms by deriving a recurrence from

f(x)

f(x-2)

First

f(20)=20^2-f(19)=306.

Subtracting through two steps gives

\begin{aligned}f(x)&=x^2-(x-1)^2+f(x-2)\\&=2x-1+f(x-2).\end{aligned}

Therefore

\begin{aligned}f(94)&=306+\sum_{j=11}^{47}(4j-1)\\&=306+4(1073)-37\\&=4561.\end{aligned}

The requested remainder is

561.