Find all polynomials \(p(x)\) such that for all real numbers \(a,b,c\neq0\) satisfying \(\frac1a+\frac1b=\frac1c\),
3 years, 6 months ago
Only polynomials of the form \(p(x)=k\cdot x\) do the job, since by taking \(b=a\) and \(c=a/2\) we have \[p(a)=2\cdot p(a/2)\]
for any non-zero real number \(a\) such that neither \(a\) or \(a/2\) belong to the zeroes of \(p(x)\).
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All that you have shown is that no real non-zero number can be the root.
You have not shown that no complex number can be the root of the polynomial, nor that 0 can be a repeated root.
@Calvin Lin : I do not agree. A polynomial is a \(C^1\) function, and there are only a finite number of \(x\) such that \(x=0,p(x)=0\) or \(p(x/2)=0\). The functional equation \(f(2x)=2f(x)\) gives that the derivative of \(f\) is constant, and the only value that \(p\) can take in zero is zero, hence \(p(x)=k\cdot x\).
This is now a more complete argument. You did not bring in the calculus aspect (that the derivative is constant) in your initial proof. Can you explain why " \( f(2x) = 2f(x) \) gives that the derivative of \(f\) is constant"?
Note that this approach doesn't require that "there are only a finite number of \(x\) such that \( x = 0 \), which was the main gist of your initial statement. All that you had initially, was "if \(a\) is a non-zero real number which is the root of the polynomial, then there are infinitely many roots, which means \( p(a) \) is a constant, which gives a contradiction. Hence there are no non-zero real roots."
That's right, \(f(x)\) can be (and, indeed, is) zero only for \(x=0\). In order to avoid calculus arguments, we can just notice that under this assumptions \(q(x)=f(x)/x\) is a polynomial satisfying \(q(2x) = q(x)\) for any \(x\neq 0\), hence \(q(x)-q(1)\) has an infinite number of roots, and \(q(x)\) is a constant polynomial.
The calculus argument is more or less the same, by deriving \( f(2x)=2f(x)\) we get \(f'(2x)=f'(x)\).