Senderov V.A.

Find all such natural $n$ that for some non-zero real numbers $a, b, c, d$ the polynomial $(a x+b)^{1000}-(c x+d)^{1000}$, after expanding the brackets and combining all like terms, has exactly $n$ non-zero coefficients.
Clearly, there exist polynomials with 1001 and 1000 non-zero coefficients (for example, $(2 x + 2)^{1000} - (x + 1)^{1000}$ and $\left.(2 x + 1)^{1000} - (x + 1)^{1000}\right)$.

Suppose that in our polynomial there are two coefficients equal to zero - at $x^{i}$ and $x^{j}$ ($i > j$). Then $a^{i} b^{1000-i} = c^{i} d^{1000-i}$,

$a^{j} b^{1000-j} = c^{j} d^{1000-j}$, which means $\left(\frac{a d}{b c}\right)^{i} = \left(\frac{b}{d}\right)^{1000} = \left(\frac{a d}{b c}\right)^{j}$. Hence, $\left|\frac{a d}{b c}\right| = 1$, $\left|\frac{d}{b}\right| = 1$, $\left|\frac{a}{c}\right| = 1$.

By replacing $a x + b$ with $(-a) x + (-b)$, our polynomial does not change. Therefore, we can assume that $a = c$. Then, if $b = d$, the resulting polynomial is zero, and if $d = -b$, in the obtained polynomial $(a x + b)^{1000} - (a x - b)^{1000}$, exactly the coefficients of even powers of $x$ are zero, resulting in 500 non-zero coefficients.

## Answer

$n=500,1000,1001$.