Problematic function

Let $y={ x }^{ 2015 }$ for some real numbers $x$ and $y$ .

So, according to the given function:

$\quad \quad f\left( { x }^{ 2015 }+1 \right) ={ x }^{ 4030 }+{ x }^{ 2015 }+1\\ \Rightarrow f\left( y+1 \right) ={ y }^{ 2 }+y+1$

Now, ${ x }^{ 2015 }-1$ can be written as $({ x }^{ 2015 }-2)+1$

So, $\Rightarrow f(({ x }^{ 2015 }-2)+1)\\ \Rightarrow f\left( (y-2)+1 \right) \quad =\quad { (y-2) }^{ 2 }+(y-2)+1\\ \Rightarrow f\left( y-1 \right) \quad =\quad ({ y }^{ 2 }+4-4y)+(y-2)+1\\ \Rightarrow f\left( y-1 \right)\quad =\quad { y }^{ 2 }+3-3y\\ \Rightarrow f\left( { x }^{ 2015 }-1 \right)\quad =\quad { x }^{ 4030 }-3({ x }^{ 2015 })+3$

Hence, the coefficients are $1, -3, +3$ , whose sum $= 1$

Cheers!:)

We need to find sum of coefficients of $f(x^{2015} - 1)$ . We get the sum of coefficients when $x = 1$ , therefore we must find $f(1^{2015} - 1) = f(0)$ .

To find $f(0)$ , we substitute $x = -1$ in the given equation. We get sum of coefficients $f(0) = (-1)^{4030} + (-1)^{2015} + 1 = \boxed{1}$ $_\square$

$f(x^{2015} + 1) = x^{4030} +x^{2015} + 1 \\ = x^{4030} +2x^{2015} +1 -(x^{2015} + 1) +1 \\ = (x^{2015}+1)^2 - (x^{2015}+1)+1 \\ \therefore f(x) = x^2-x+1$

So $f(x^{2015}-1) = (x^{2015}-1)^2 - (x^{2015}-1)+1 \\ = x^{4030} -3x^{2015}+3$

Hence required answer is $1-3+3 = 1$

We may rewrite $f$ as $f(x^{2015} + 1) = x^{2015} (x^{2015} + 1) + 1$ . Setting $x^{2015} + 1 = u$ we have $f(u) = u^{2} - u + 1$ . Letting $u = x^{2015} - 1$ gives us $f(x^{2015} - 1) = x^{2030} - 3x^{2015} + 3$ .