An algebra problem by Anand Raj

Algebra Level 2

Written on a blackboard is the polynomial ${ x }^{ 2 }+x+2014$ Calvin and Hobbes take turns alternatively (starting with Calvin) in the following game. During his turn, Calvin should either increase or decrease the coefficient of x by 1. And during his turn, Hobbes should either increase or decrease the constant coefficient by 1. If Calvin wins if at any point of time the polynomial on the blackboard at that instant has integer roots, then Who Has A Winning Strategy?

The Game Is Endless Hobbes Can't Say Calvin

2 solutions

Aditya Raut
Jul 1, 2014

This problem is actually the 4th problem of INMO 2014, I know because I had given this exam.

$\color{#D61F06}{\boxed{Calvin}}$ has the $\color{#20A900}{Winning \ Strategy}$ .

I think you forget to look down a little bit!!!!

Anand Raj - 6 years, 11 months ago

Haha sorry for that @Anand Raj ... really forgot to look down a little bit...

$\Huge{\color{#3D99F6}{S}} \color{#D61F06}{\Huge{O}} \color{#20A900}{\Huge{R}} \color{#69047E}{\Huge{R}} \color{limegreen}{\Huge{Y}}$

Aditya Raut - 6 years, 11 months ago

An interesting follow-up question is, could Calvin still win if the condition of victory is modified to exclude any polynomial where 2 or -2 is a root?

Peter Byers - 4 years, 4 months ago

R.L. Bhat
Jul 2, 2014

If b be the coefficient of x, to get integer roots, we must have b^2 - 4*2014 must be a perfect square. To get the real roots, b^2 should be more than 8000. Try 90, 91 and so on. 91 satisfies the condition. 91 is odd. So Calvin wins.

But Hobbes can increase or decrease the constant coefficient by one each turn. So he can't change it to some value other than 2014,, right before Calvin reaches 91?

Casper Putz - 6 years, 11 months ago

An algebra problem by Anand Raj

2 solutions

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