'Palindrome' rooted polynomial.

Algebra Level 5

$x^8 + 2x^6 + 2x^5 + 2x^4 + 2x^3 + 2x^2 + x = 6039306$

Find sum of all the integer solutions to the equation above.

Also try Taxicab Rooted Polynomial .

The answer is 7.

1 solution

Nihar Mahajan
Mar 4, 2015

$x^8 + 2x^6 + 2x^5 + 2x^4 + 2x^3 + 2x^2 + x = 6039306$

$(x^4+x^2+x)(x^4+x^2+x+1) = 6039306$

Let $x^4 + x^2 + x = a$

$a(a+1)=6039306$

$a^2 + a - 6039306=0$

$(a-2457)(a+2458) = 0$

When $a = -2458$ , it yields complex roots.

When $a = 2457 \Rightarrow x^4 + x^2 + x - 2457=0$

$\Rightarrow (x-7)(x^3 + 7x^2 + 50x + 351)=0$

$(x^3 + 7x^2 + 50x + 351)=0$ yields 2 complex and one pathetic real root.

Hence , the only integer solution to the equation is $\huge\boxed{7}$

Moderator note:

Poor choice of words. What does "one pathetic real root" mean? It's easy (though a little tedious) to prove that the cubic equation in your penultimate line has no integer roots using Rational Root Theorem. And you need to explain how you factorize such a large number ( $6039306$ ) that quickly!

Nice set of problems by you Nihar. I appreciate your capabilities!! Keep it up...

Sanjeet Raria - 6 years, 3 months ago

Thank you very much! :) :)

Nihar Mahajan - 6 years, 3 months ago

Your original equation should be $x^8 + 2x^6 + 2x^5 + 2x^4 + 2x^3 + 2x^2 + x = 6039306.$ (The coefficient of $x^2$ should be 2.)

Jon Haussmann - 6 years, 3 months ago

Yeah! I am extremely sorry about that.... typo mistake.Fixed.