Singapore Maths Olympiad

If it is known that the roots of the equation are all integers, how many distinct roots does the equation have?

x⁵ + 3x⁴ – 4044118x³ – 12132362x² – 12132363x - 2011² = 0

Singapore maths Olympiad

answer:

Using Descartes rule of signs

if p(x) is the polynomial, then number of +ve roots equal to the number of variations of sign of P(x) or less than by an even number.

P(x)= x⁵ + 3x⁴ – 4044118x³ – 12132362x² – 12132363x - 2011²

Sign change = 1 

so, number of +ve roots = 1

Similarly

if p(x) is the polynomial, then number of -ve roots equal to the number of variations of sign of P(-x) or less than by an even number.

P(-x) = -x⁵ + 3x⁴ + 4044118x³ – 12132362x² + 12132363x - 2011²

So, number of sign change = 4

So, the number of negative roots = 4 or 2. But since the number of + roots is 1 and the total number of roots are 5, therefore the number of -ve roots has to be 4.

Now since the roots are all integers, therefore the roots have to be divisors of 2011²

Since 2011 is prime, so the roots can be ± 1, ± 2011, ± 2011²

But since product of all roots =  -2011²

This is only possible if the roots are -1, -1, -1, -2011, + 2011

So, there are 3 distinct roots.