Indian National Mathematic Olympiad 2017 Q2

August 14, 2023

1. Problem

Suppose ${n >= 0}$ is an integer and all the roots of ${x^3 + \alpha x + 4 - (2 * 2016^n) = 0}$ are integers. Find all possible values of ${\alpha}$

2. Solution

Let’s say a,b,c are integers and roots of

$\displaystyle x^3 + \alpha x + 4 - (2 * 2016^n) = 0 \ \ \ \ \ (1)$

We can write this polynomial as

$\displaystyle (x-a)(x-b)(x-c) = 0$

$\displaystyle x^3- (a+b+c)x^2 + (ab+bc+ca)x -abc =0 \ \ \ \ \ (2)$

Comparing (1) and (2)

$\displaystyle a+b+c =0 \ \ \ \ \ (3)$

$\displaystyle ab+bc+ca = \alpha \ \ \ \ \ (4)$

$\displaystyle -abc = 4 - (2 * 2016^n) \ \ \ \ \ (5)$

Using (3) and (5)

$\displaystyle ab(a+b) = 4 - (2 * 2016^n) \ \ \ \ \ (6)$

Let’s say ${n=0}$ ,

from (6),

$\displaystyle ab(a+b) = 2$

since a and b are integer values can only range from -2 and 2. We can easily eliminate 0 as it is not the factor of 2.

ab(a+b)		b
ab(a+b)		-2	-1	1	2
a	-2	-16	-6	2	0
	-1	-6	-2	0	-2
	1	2	0	2	6
	2	0	-2	6	16

From above table, we can see roots which satisfies equation 6 are 1,1,-2. Using these roots in (4)

$\displaystyle \alpha = 1 -2 -2 = -3$

Now, let’s say ${n > 0}$ , from (6),

$\displaystyle ab(a+b) \equiv 4 \equiv 1~(mod~ 3)$

So, a and b cannot be divisible by 3

ab(a+b)		b
ab(a+b)		1	2
a	1	2	0
a	2	0	1

So,

$\displaystyle a \equiv b \equiv 2 ~(mod~3) \ \ \ \ \ (7)$

from (6),

$\displaystyle ab(a+b) \equiv 4~ (mod ~9)$

and from (7)

$\displaystyle a \equiv 2,~5~ or~ 8 ~(mod ~ 9)$

$\displaystyle b \equiv 2,~5~ or~ 8 ~(mod ~ 9)$

ab(a+b)		b
ab(a+b)		2	5	8
a	2	7	7	7
	5	7	7	7
	8	7	7	7

None of the above combination will satisfy ${ab(a+b) \equiv 4 ~(mod ~9) }$ .

So, for ${n > 0}$ there is no solution for which equation (1) has all integer roots.

So, the only solution for the equation is when ${\mathbf{\alpha = -3}}$ and roots are 1,1,-2.

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Indian National Mathematic Olympiad 2017 Q2

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