Source: 2022 Bulgarian Spring Math Competition, Problem 9.1
Tags: algebra, quadratic trinomial, quadratics, function
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Let $f(x)$ be a quadratic function with integer coefficients. If we know that $f(0)$, $f(3)$ and $f(4)$ are all different and elements of the set $\{2, 20, 202, 2022\}$, determine all possible values of $f(1)$.
$f(x) = ax^2+bx+c$
$f(0) = c $
$f(3) = 9a+3b+c $
$f(4) = 16a+4b+c$
Case $f(0)=2$
$f(3) = 9a+3b+2 \equiv 20 \mod 3 $ ( 2 cannot be used since f(0)=2)
$9a+3b=18,3a+b=6$
$f(4) = 16a+4b+2 \equiv (202,2022) \mod 4$
subCase (f(4)=202)
$16a+4b=200$
$4a+b = 50$
$a+b+c = 44+6-44*3+2 = 8-44*2 = -80$
subCase (f(4) 2022)
$16a+4b=2020$
$4a+b = 505$
$3a+b=6$
$a+b+c=-990$
case $f(0)=20$
$f(4) = 16a+4b+20 \equiv 0 \mod 4$
forcing f(4) = 20 contradiction
case $f(0) = 202$
$f(3) = 9a+3b+202 \equiv 1 \mod 3$
forcing $f(3) = 202$ contradiction
case $f(0) = 2022$
$f(3) = 9a+3b+2022 \equiv 0 \mod 3 $
forcing $f(3) = 2022$ contradiction