Parity - "A Lethal Weapon"

This note is to show how \(‘PARITY’\) can be used very effectively to solve olympaid problems .

$'PARITY’$ means that $odd+odd=even,odd+even=odd,even+even=even,odd(odd)=odd,odd(even)=even,even(even)=even$

To explain its significance a example of question of $RMO 2016$

QUESTION : $f(x)=x^{3}-(k-3)x^{2}-11x+(4k-8)$ Find all integers $k$ such that roots of $f(x)$ are integers.

Proof : let $a,b,c$ are integral roots of $f(x)$

This implies $a+b+c=(k-3),ab+bc+ac=-11,abc=4(2-k)$

Since $a,b,c$ are integers and $abc=4(2-k)$ therefore atleast one of them is even or $k=2$ but $k=2$ doesn’t give integral roots but $ab+bc+ac=-11$ therefore at a time 2 or 3 of ${a,b,c}$ can’t be even .

Therefore only one of them is even .

Let $a$ is even .

Also $a+b+c=even;k-3=even;k=odd$

$f(x)=(x-a)(x-b)(x-c)$ , this implies for $x$ be an integer $f(x)$ is also an integer . Now put $x=2$(because at $x=2$ the $k$ term diappeas in $f(x)$) in $f(x)$ this gives $f(2)=(2-a)(2-b)(2-c)=-10$ . Now since $a$ is even therefore $(2-a)={-2,2}$ where $2-a=-2$ is the essential case as $2-a=2$ makes $k=even$.

So $a=4$ is a root of $f(x)$ . after putting values of $x=4$ in $f(x)$ we get $k=5,b=1,c=-3$ which matches our condition that ${k,b,c}=odd$ .