Prove that if $f(x) = Ax^2 + Bx + C$ is an integer whenever $x$ is an integer, then $2A$, $A+B$ and $C$ are also integers.

Question

Prove that if $f(x) = Ax^2 + Bx + C$ is an integer whenever $x$ is an integer, then $2A$, $A+B$ and $C$ are also integers.

I've tried a lot to do it, but can't get it exactly right.

Answer 1

We have

$f(x) = Ax^2 + Bx + C; \tag{1}$

set $x = 0$, an integer. Then we have

$f(0) = C, \tag{2}$

so $C$ is an integer. Set $x = 1$; then

$f(1) - C = A + B, \tag{3}$

showing $A + B$ is an integer. Set $x = -1$; then

$f(-1) - C = A - B \tag{4}$

is also an integer. Since

$2A = (A + B) + (A - B), \tag{4}$

we see that $2A$ is an integer as well. Noting

$2B = (A + B) - (A - B) \tag{5}$

we also see $2B$ is an integer, "for free". It appears to me that lab bhattacharjee was onto the right idea, though I did work this out for myself, concurrently as it were. Cute, I must say . . .

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

Answer 2

Hint $\displaystyle\,\ f(x) = ax^2\!+bc+c\, =\, 2a\, \dfrac{x(x\!-\!1)}2 + (a\!+\!b)\, x + c \,=\, 2a {x \choose 2} + (a\!+\!b) {x\choose 1} + c{x\choose 0}$

Lemma $\ $ If $\displaystyle\ f(x)\, =\, c_2 {x \choose 2} + c_1 {x \choose 1} + c_0{x\choose 0}\ $ then $\ f(0),f(1),f(2)\in\Bbb Z\,\Rightarrow\, c_i\in \Bbb Z$

Proof $\ f(0) = c_0\in \Bbb Z\,$ so $\, f(1) = c_0\!+\!c_1\in\Bbb Z\Rightarrow c_1\in\Bbb Z\,$ so $\, f(2) = c_0\!+\!2c_1\! +\! c_2\in\Bbb Z\,\Rightarrow\,c_2\in\Bbb Z$

Remark $\ $ The analogous result (and its converse) is true for arbitrary degree, which leads to a famous result of Polya and Ostrowski on integer-valued polynomials