Theory of Equations 1 Question 46
47. Let $f(x)=A x^{2}+B x+C$ where, $A, B, C$ are real numbers. prove that if $f(x)$ is an integer whenever $x$ is an integer, then the numbers $2 A, A+B$ and $C$ are all integers. Conversely, prove that if the numbers $2 A, A+B$ and $C$ are all integers, then $f(x)$ is an integer whenever $x$ is an integer.
$(1998,3$ M)
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Solution:
- Suppose $f(x)=A x^{2}+B x+C$ is an integer, whenever $x$ is an integer.
$\therefore \quad f(0), f(1), f(-1)$ are integers.
$\Rightarrow \quad C, A+B+C, A-B+C$ are integers.
$\Rightarrow \quad C, A+B, A-B$ are integers.
$\Rightarrow \quad C, A+B,(A+B)-(A-B)=2 A$ are integers.
Conversely, suppose $2 A, A+B$ and $C$ are integers.
Let $n$ be any integer. We have,
$f(n)=A n^{2}+B n+C=2 A \frac{n(n-1)}{2}+(A+B) n+C$
Since, $n$ is an integer, $n(n-1) / 2$ is an integer. Also, $2 A, A+B$ and $C$ are integers.
We get $f(n)$ is an integer for all integer $n$.
