Parity - "A Lethal Weapon"

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

PARITY'PARITY’ means that odd+odd=even,odd+even=odd,even+even=even,odd(odd)=odd,odd(even)=even,even(even)=evenodd+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 RMO2016RMO 2016

QUESTION : f(x)=x3(k3)x211x+(4k8)f(x)=x^{3}-(k-3)x^{2}-11x+(4k-8) Find all integers kk such that roots of f(x)f(x) are integers.

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

This implies a+b+c=(k3),ab+bc+ac=11,abc=4(2k)a+b+c=(k-3),ab+bc+ac=-11,abc=4(2-k)

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

Therefore only one of them is even .

Let aa is even .

Also a+b+c=even;k3=even;k=odda+b+c=even;k-3=even;k=odd

f(x)=(xa)(xb)(xc)f(x)=(x-a)(x-b)(x-c) , this implies for xx be an integer f(x)f(x) is also an integer . Now put x=2x=2(because at x=2x=2 the kk term diappeas in f(x)f(x)) in f(x)f(x) this gives f(2)=(2a)(2b)(2c)=10f(2)=(2-a)(2-b)(2-c)=-10 . Now since aa is even therefore (2a)=2,2(2-a)={-2,2} where 2a=22-a=-2 is the essential case as 2a=22-a=2 makes k=evenk=even.

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

Note by Shivam Jadhav
4 years ago

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Awesome use of parity!

Harsh Shrivastava - 4 years ago

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Great solution!

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

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ab+bc+ca=-11 not satisfied

Shivam Jadhav - 4 years ago

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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.

Niranjan Khanderia - 3 years, 5 months ago

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