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 \({k,b,c}=odd\) .

## Comments

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TopNewestAwesome use of parity! – Harsh Shrivastava · 5 months, 1 week ago

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Why didn't you consider the case 2-a=|10|? – Svatejas Shivakumar · 5 months, 1 week ago

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– Shivam Jadhav · 5 months, 1 week ago

ab+bc+ca=-11 not satisfiedLog in to reply

Great solution! – Svatejas Shivakumar · 5 months, 1 week ago

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@Harsh Shrivastava @Svatejas Shivakumar @Nihar Mahajan – Shivam Jadhav · 5 months, 1 week ago

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@Calvin Lin @Daniel Liu @Pi Han Goh @Chinmay Sangawadekar – Shivam Jadhav · 5 months, 1 week ago

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