SILK ROAD MATHEMATICS COMPETITION XVII

Answer: $f(x)=x$ for all real $x$. From the first equation we have by induction that $$f(x+n)=f(x)+n(1)$$ for every real $x$ and integer $n$. Since $x^4-x^2=(x^2-\frac{1}{2}{)}^2-\frac{1}{4}$, for every $y\ge -\frac{1}{4}$ there is $x$ such that $y=x^4-x^2$, implying $$f(y)=f(x^4-x^2)=f(x{)}^4-f(x{)}^2={\left(f(x{)}^2-\frac{1}{2}\right)}^2-\frac{1}{4}\ge -\frac{1}{4}.$$

For every $x$ we have $x=\{x\}+[x]$ and $0\le \{x\}<1$ (in particular, $\{x\}>-\frac{1}{4}$ and $[x]>x-1$), implying $$f(x)=f(\{x\})+[x]>-\frac{1}{4}+(x-1)>x-2.(2)$$ Suppose there is $x$ such that $t=f(x)-x\ne 0$. Then for every integer $n$ we have $$f(x+n{)}^4-f(x+n{)}^2=f((x+n{)}^4-(x+n{)}^2),$$ giving $$(x+n+t{)}^4-(x+n+t{)}^2>(x+n{)}^4-(x+n{)}^2-2,$$ and thus $$4t(x+n{)}^3+6t^2(x+n{)}^2+2t(2t^2-1)(x+n)+t^4-t^2+2>0(3)$$ for every integer $n$, which is impossible as if $t>0$, then for $n\to -\infty$ we get a contradiction. If $t<0$, then for $n\to \infty$ we again get a contradiction. Therefore, $f(x)=x$ for all real $x$ which is easy to check works for both given equations.