# 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 $${k,b,c}=odd$$ .

1 year, 6 months ago

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Awesome use of parity!

- 1 year, 6 months ago

Yes. Parity is of great help in many cases. It needs a little practice before we can make great use of it. I had used it in silution of quadratic equations. Thank you for the notes.

- 11 months, 3 weeks ago

Why didn't you consider the case 2-a=|10|?

- 1 year, 6 months ago

ab+bc+ca=-11 not satisfied

- 1 year, 6 months ago

Great solution!

- 1 year, 6 months ago

- 1 year, 6 months ago