Why Are There So Many Powers?

Let's change $x^4f(x)^4+x^2=3x^3f(x)^2$ into a more reasonable equation.

First, move $3x^3f(x)^2$ to the LHS to get $x^4f(x)^4-3x^3f(x)^2+x^2=0$ .

Then, divide by $x^2$ to get $x^2f(x)^4-3xf(x)^2+1=0$ .

Then, solving for $xf(x)^2$ using the quadratic formula, we get $xf(x)^2=\frac{3\pm\sqrt{5}}{2}$ . Substituting $x=2014$ , we get $2014f(2014)^2=\frac{3\pm\sqrt{5}}{2}$ .

Next, we divide both sides by 2014 to get $f(2014)^2=\frac{3\pm\sqrt{5}}{4028}$ .

Next, we square root both sides and get $f(2014)=\pm\sqrt{\frac{3\pm\sqrt{5}}{4028}}$ . We can throw out the negative values of $f(2014)$ since they're not going to be the largest value of $f(2014)$ . We see that $\sqrt{\frac{3-\sqrt{5}}{4028}}\approx0.01377$ and $\sqrt{\frac{3+\sqrt{5}}{4028}}\approx0.03605$ , so we take the larger value, which is $\sqrt{\frac{3+\sqrt{5}}{4028}}\approx0.03605$ , and let it be $M$ .

Lastly, we find $\lfloor1000M\rfloor$ , which is $\lfloor1000\sqrt{\frac{3+\sqrt{5}}{4028}}\rfloor \approx \lfloor36.05\rfloor = \boxed{36}$ .