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Find all polynomials $P$ with real coefficients that satisfy the condition: $P(x)$ is a rational number if and only if $x$ is a rational number.

I find that the rational coefficient polynomials are all satisfied but can not shown that they are unique :( pls help

Newbiemath
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    2 = x^2 seems to violate your hypothesis, yeah? – ZKe Aug 26 '23 at 03:29
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    Lagrange Interpolation shows that a polynomial of degree at most $n$ that takes rational values at $n+1$ distinct rationals must have rational coefficients. – Arturo Magidin Aug 26 '23 at 03:35
  • I don't think this is a duplicate of the linked question. That question certainly plays a part as a subproblem, but (unlike OP believes), it is far from the whole answer, as rational polynomes of degree > 1 aren't answers to this question. – Ingix Aug 26 '23 at 07:06

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Suppose a polynomial, $f_L$, of degree $L \geq 2$ satisfies this property. By fundamental theorem of algebra, $f_L = (x - \lambda)f_{L-1}$ for some $\lambda \in \mathbb{R}$ and some polynomial $f_{L-1}$ of degree L-1. Trivially, $L \in \mathbb{Q}$ and $f_{L-1}(x)$ must be a rational number if and only if x is a rational number. I claim that there are no polynomials of degree 2 with this property, implying that there are no polynomials of degree $\geq 2$ with this property. Suppose f(x) = $ax^2 + bx + c$. Suppose f(x) has the above property. This means that a, b and c are all rational. $f(x) = a(x + \frac{b}{2a})^2 + c - \frac{b^2}{4a}$ Let $i$ be the smallest integer $\geq 0$ such that $a^{\frac{1}{2^i}}$ is irrational. Let $x =a^{\frac{1}{2^i}}- \frac{b}{2a}$. Clearly, x is irrational and f(x) is rational. The case of polynomials with degree 1 is trivial and left as an exercise

mscj
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    I don't think that works out. First, $f_L$ may not have a real root $\lambda$ at all. Second, even if such $\lambda$ exists, $f_{L-1}$ may very well map an irrational to a rational number. For an irrational $x$, $f_{L-1}(x) = f_L(x)/(x-\lambda)$ is is a quotient between the (assumed) irrational $f_L(x)$ and the irrational $x-\lambda$. Nothing says this quotient is irrational. – Ingix Aug 26 '23 at 07:00