Squares of polynomials with all nonzero integer coefficients [duplicate]

For polynomials of one variable with only and all nonzero integer coefficients, do there exist any where the polynomial consists of more than one term, and the square of the polynomial has the same number of terms or fewer than the polynomial?

For instance (to contrast), $$(x + 1)^2 = x^2 + 2x + 1$$

The polynomial in parentheses has two terms, but the square of the polynomial has three terms.

Another instance, $$(x^2 - x + 2)^2 = x^4 - 2x^3 + 5x^2 - 4x + 4$$

The polynomial in parentheses has three terms, but the square of the polynomial has five terms.

Note: This is not a duplicate of another question. That merely requested that that [p(x)]^2 had to have fewer nonzero terms than p(x). Mine requires that p(x) 1) must have all of its terms' coefficients be nonzero and integers, and 2) the number of terms could be equal, not just fewer.

marked as duplicate by gt6989b, Robert Israel, GNUSupporter 8964民主女神 地下教會, Leucippus, Vinyl_cape_jawaMar 12 at 1:19

• Note that Jack D'Aurizio's answer to this question gives a polynomial $p$ such that $p$ and $p^2$ have the same number of terms. – saulspatz Mar 11 at 20:08

Yes, and apparently the minimal examples occur in degree $$12$$. Coppersmith and Davenport found a class of examples: Define $$p_{\alpha}(x) := (125 x^6 + 50 x^5 - 10 x^4 + 4 x^3 - 2 x^2 + 2 x + 1) (\alpha x^6 + 1) .$$ Expanding shows that none of the $$13$$ coefficients are $$0$$ provided $$\alpha \neq 0, -125$$, but for eight particular values $$\alpha$$, including $$\alpha = -110$$, $$p_\alpha(x)^2$$ has precisely 12 nonzero coefficients.