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Define the term “closer to” for polynomials in the intuitive way: $B(z)$ is closer to $A(z)$ then $C(z)$ if the degree of $\lvert A(z) \pm B(z) \rvert$ is smaller than the degree of $\lvert A(z) \pm C(z) \rvert$, or the coefficients are smaller in the case the degrees are equal.

Now let's say I have an arbitrary polynomial, e.g. $$ A(z) := 4z^4-28z^3+60z^2-43z+8. $$ Is there an algorithm to determine the closest square [or triangular or cubic, etc.] polynomial to $A(z)$? For example, $$ 4z^4-28z^3+60z^2-43z+8 = (2z^2-7z+3)^2-(z^2+z+1), $$ and so $(2z^2-7z+3)^2$ is the closest polynomial square to $A(z)$. But I know this only because I constructed the example in the first place!

EDIT: Here's a new concrete example. I'm trying to show that $$3x^4+3x^2+1$$ can never be a square. So how can I calculate the nearest square above and below?

Kieren MacMillan
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    Chebychev polynomials are your (and my) friends. – marty cohen May 08 '14 at 02:45
  • A-ha! Thank you for that direction. Quickly scanning http://en.wikipedia.org/wiki/Approximation_theory#Optimal_polynomials shows me the direction I need to search. – Kieren MacMillan May 08 '14 at 02:53
  • If anyone has a link to a short example or tutorial on exactly how to use Chebychev polynomials to "divide" a polynomial into its largest square and remainder, I'd appreciate it. – Kieren MacMillan May 08 '14 at 10:41
  • @martycohen: Do you have a link or short example on how to use Chebyshev polynomials to show that $(2z^2-7z+3)^2$ is the [polynomial] square closest to $4z^4-28z^3+60z^2-43z+8$? – Kieren MacMillan Jul 15 '21 at 21:10
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    This would not involve Chebychev polynomials. Write your polynomial as $4z^4(1-7/z+15/z^2...)$ and then use the expansion $(1+x)^{1/2}=1+x/2-x^2/8+...$ with $x$ being the terms in $1/z$. See, for example, https://en.wikipedia.org/wiki/Binomial_theorem#Newton's_generalized_binomial_theorem – marty cohen Jul 16 '21 at 16:26
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    Wolfram Alpha, when given $\sqrt(4 z^4 - 28 z^3 + 60 z^2 - 43 z + 8)$, returns, as part of its answer, $2 z^2 - 7 z + 11/4 - 9/(8 z) + O((1/z)^2)$ as the expansion at $z=\infty$. – marty cohen Jul 16 '21 at 16:30
  • @martycohen: Please make that expansion comment into an answer, so I can accept it. Thanks! – Kieren MacMillan Jul 16 '21 at 17:06

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(at OP's request)

Wolram Alpha, when given $\sqrt{4 z^4 - 28 z^3 + 60 z^2 - 43 z + 8}$, returns, as part of its answer, $2^2−7+11/4−9/(8)+((1/)^2)$ as the expansion at =∞.

You then have to try integer values for the 11/4 to see which is closest.

For some reason, this does not work when given $(4 z^4 - 28 z^3 + 60 z^2 - 43 z + 8)^{1/2} $.

marty cohen
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