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does there exist a positive integer so that is a perfect square polynomial
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Since f(x) must be a perfect square polynomial, then all its roots must be multiple roots. In the same time, all those roots must be roots of f'(x) with a well-known consequency to their multiplicities.
Degree of f'(x) is 2n-1, and if f'(x) contained two distinct roots, then a degree of f(x) would be bigger than 2n -- a contradiction.
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Therefore f'(x) must be of a form with the unique real root (consider complex conjugates to realize that the root must be a real one), and comparing the absolute terms, we have the only candidate .
It is easy to see (verify by yourself) that since the absolute term of is , then it is equal to 1 only if n=1.
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Thanks musixx for solving my question also thanks to all members to show interest on my questions.
and byk7 you are saying right I am from India.
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↑ stuart clark: Can I ask you, too? How/when did you find this website?
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Let . Then
Prove that for any .
The sequence is decreasing therefore
However, values of a perfect square polynomial can be only nonnegative real numbers. Hence cannot be a perfect square polynomial.
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Actually I have got this site 2 months ago with the help of google search.
Thanks povel for explanation.
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