Does there exist a polynomial $f(x)$ with real coefficients such that $f(x)^2$ has fewer nonzero coefficients than $f(x)$?

Hint: What happens if you consider a truncated Taylor series for $\sqrt{1-x}$ ? For instance:

$$\left(1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5 x^4}{128}\right)^2 = 1-x+\frac{7 x^5}{128}+\frac{7 x^6}{512}+\frac{5 x^7}{1024}+\frac{25 x^8}{16384}$$ with about the same number of non-zero coefficients. Are you able to adjust a couple of coefficients in the LHS in order to prove your claim? For instance: $$\left(1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+ \frac{x^4}{64}\right)^2=1-x+\frac{7 x^4}{64}-\frac{x^7}{512}+\frac{x^8}{4096}.$$ However, in order to find a polynomial whose square has fewer non-zero terms than the original polynomial, you have to consider polynomials with degree at least $12$.