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Let $P(x)$ be a polynomial, with leading coefficient equal to $1$, of degree $1991$ (with integral coefficients). Prove that the polynomial $Q(x) = P(x)^2−9$ has at most $1995$ distinct integral roots.

So obviously my first step was to express $Q(x)$ as $Q(x)=(P(x)-3)(P(x)+3)$. $P(x)$ can have at most $1991$ roots and its leading coefficient is $1$, hence we can also write $P(x)$ as $P(x)=(x-x_1)\cdot(x-x_2)\cdot{\dots}\cdot(x-x_{1991})$. I also tried to define polynomials $F_1(x)=P(x)-3$, $F_2(x)=P(x)+3$ so then $Q(x)=F_1(x)F_2(x)$ where degree of $F_1(x), F_2(x)$ is also equal at most to $1991$, but I can't see how it would lead to $1995$ possible pairwise distinct values of $x$ for which $Q(x)=0$. Any help what to do next will be appreciated, thanks!

J. W. Tanner
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krxl
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  • Hint: A well known fact of polynomials with integer coefficients is that $ a - b \mid f(a) - f(b)$. How can we apply this here? – Calvin Lin Dec 30 '24 at 17:13
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    See this $P^2(x) - 1)$ has at most $\deg P + 2$ roots as a close-duplicate. $\quad$ The condition on monic is not required. $\quad$ In fact, we can even show that it has at most 1991 distinct integral roots. – Calvin Lin Dec 30 '24 at 17:20
  • @CalvinLin: I have never come across your "well known fact" before. Can you provide a reference please. – Rob Arthan Dec 30 '24 at 21:31
  • @RobArthan Clearly $ a - b \mid a^b - b^n$ so $ a-b \mid \sum f_i( a^n - b^n ) = f(a) - f(b)$. See this MSE post for more details if needed. – Calvin Lin Dec 30 '24 at 22:00
  • @CalvinLin: thanks for that. – Rob Arthan Dec 30 '24 at 22:08
  • @CalvinLin can only see how to limit numbers to $2d$ (here $d=1991$) but I don't get why $d-2$ roots of $F_1(x), F_2(x)$ overlap. I tried to use your formula but I don't see any polynomial that we know two values of. The way how to aply the close-duplicate you sent seems beyond my understanding sadly. – krxl Dec 31 '24 at 11:36

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