How to prove that every polynomial with real coefficients is the sum of three polynomials raised to the 3rd degree? Formally the statement is:
$\forall f\in\mathbb{R}[x]\quad \exists g,h,p\in\mathbb{R}[x]\quad f=g^3+h^3+p^3$
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Glinka
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5I'm confused by this question. Surely a linear function or a quadratic for $f$ is a simple counter example. – Ian Miller Oct 30 '16 at 08:24
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18@IanMiller No. $3x^2+3x=(x+1)^3+(-x)^3+(-1)^3.$ – mfl Oct 30 '16 at 08:29
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Oh right. My bad. – Ian Miller Oct 30 '16 at 09:25
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2What is this, a contest question or something? – Jack M Oct 30 '16 at 23:11
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@JackM, yes, it was offered on a contest – Glinka Oct 30 '16 at 23:45
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1@Glinka what contest? – frog1944 Nov 12 '16 at 01:36
1 Answers
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We have that the following identity holds $$(x+1)^3+2(-x)^3+(x-1)^3=6x.$$ Hence $$\left(\frac{f(x)+1}{6^{1/3}}\right)^{3}+\left(\frac{-f(x)}{3^{1/3}}\right)^{3}+ \left(\frac{f(x)-1}{6^{1/3}}\right)^{3}=f(x).$$
Robert Z
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22Great answer. This appears to be extendable to all powers as:
$$\sum_{i=0}^{n-1}\left(x-\frac{n-1}{2}+i\right)^n\cdot(-1)^i\cdot{n-1\choose i}=n!\cdot x$$
– Ian Miller Oct 30 '16 at 10:08 -
3@IanMiller For $n = 2$ you get $-2x$ on the left. I think the right side should be $(-1)^{n-1} n! \cdot x$ – Jay Oct 31 '16 at 00:15
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5Over the rationals: I wonder for what $f\in\mathbb{Q}[x]$ there exist $g,h,p\in\mathbb{Q}[x]$ such that $f=g^3+h^3+p^3$ (since the cube root of six is clearly irrational). – Jeppe Stig Nielsen Oct 31 '16 at 10:20
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Does this include for cubed perfect numbers as well? How every perfect number cubed can be written as the sum of three cubes? I believe this has been conjectured – Mr Pie Oct 21 '17 at 12:29