p(x)=1 and p(x)=3 have integer solutions, can p(x)=2 have 2?

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p(x)=1 and p(x)=3 have integer solutions, can p(x)=2 have 2?

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#1 h Dec 22, 2011, 3:00 PM • 2 Y

Y by Adventure10, Mango247

Let $p(x)$ be a polynomial with integer coefficients such that both equations $p(x)=1$ and $p(x)=3$ have integer solutions. Can the equation $p(x)=2$ have two different integer solutions?

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#2 h Dec 22, 2011, 4:16 PM • 2 Y

Y by Adventure10, Mango247

Suppose $p(x)=2$ has at least 2 distinct integer roots $y_1,y_2$ . Then $x$ is a integer root of equation $p(x)=1$ , $z$ is a integer root of equation $p(x)=3$ , $x\neq z$ . It's easy to see that $y_1-x, y_2-x, y_1-z,y_2-z | 1$ And there are just 2 cases, and both also reach a contradiction!!. So Answer is "No"

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