xeroxia wrote: $f$ is a function defined on integers and satisfies $f(x)+f(x+3)=x^2$ for every integer $x$. If $f(19)=94$, then calculate $f(94)$. It is too easy to be an olympiad exercise. $ f(22)=19^2-94 $ $ f(25)=22^2-19^2+94 $ So it is easy to see that $ f(94)=91^2-88^2+85^2-........+19^2-94 $ $ =3(91+88+85+......+22)+19^2-94 $ $ =3*12*113+361-94 $ An easy calculation does the rest.

$f(94) = 91^2 - f(91) $ $ = 91^2 - 88^2 + f(88)$ $= 91^2 - 88^2 + 85^2 - f(85)$ . . . $= 91^2 - 88^2 + 85^2 - 82^2 + ... + 25^2 - 22^2 + 19^2 - f(19)$ $= \sum_{k=4}^{15} [(6k+1)^2 - (6k-2)^2] + 19^2 - 94$ $= \sum_{k=4}^{15} 3(12k-1) + 19^2 - 94$ $= 36 \sum_{k=4}^{15} k - \sum_{k=4}^{15} 3 + 19^2 - 94$ $= 36 \frac{(12)(19)}{2} - 3(12) + 19^2 - 94 = \boxed{4335}$

f(x)+f(x+3)=x^2 implies (with the approach f(x)=ax^2+bx+c) - as you can easily check - f(x)=x(x-3)/2, which fulfills the functional equation, but not the condition f(19)=94. In comparison with the solutions above the number 94 corresponds to f(19)=152. As a consequence, we get f(94)=94*91/2+(152-94)=47*91+58=4277+58=4335.

Casas-Alvero wrote: f(x)+f(x+3)=x^2 implies (with the approach f(x)=ax^2+bx+c) - as you can easily check - f(x)=x(x-3)/2, which fulfills the functional equation, but not the condition f(19)=94. In comparison with the solutions above the number 94 corresponds to f(19)=152. As a consequence, we get f(94)=94*91/2+(152-94)=47*91+58=4277+58=4335. $f$ is not a polynomial.