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Can any one give me example of: rational functions $f, g$ and $h$ with rational coefficients such that

$$(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$$

Also, if anyone knows a procedure for constructing such examples, I would be happy to learn them as well.

Added: Can we find rational functions $f$ and $g$ such that $(f(x))^3+(g(x))^3=x$

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I don't know the answer, but by clearing denominators, you reduce this to $f(x)^3+g(x)^3+h(x)^3=x q(x)^3$, for polynomials with integer coefficients. This may be easier to think about. Setting $x=1$, you get an integer solution of $a^3+b^3+c^3=d^3$. This paper is relevant then. Also, by setting $x=0$, you get Fermat's equation for exponent 3, which only has trivial solutions. In particular, $x$ divides one of $f$, $g$, $h$. Not that it helps... –  lhf May 5 '11 at 13:00
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Another observation is that (using lhf's notation) we must have all coefficients of $q(x)$ divisible by $3$. This is obtained by modulo 3 on both side, so we get $(f+g+h)^3=xq^3$ in the polynomial ring $\mathbb{F}_3[x]$, but $x$ is not a perfect cube, so this happens only when both sides are zero. –  Jiangwei Xue May 5 '11 at 13:50
    
Start with $f(x)^3+g(x)^3+h(x)^3=x q(x)^3$. Suppose $q(x)$ has degree $k$. Then the RHS is asymptotic to $cx^{3k+1}$ as $x \to \infty$. But the LHS is asymptotic to $dx^{3m}$ unless there is cancellation of the highest term... By exponent 3 of Fermat, that means two of the terms have higher degree than the third, and their highest terms are negatives. –  GEdgar May 5 '11 at 13:57
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4 Answers

$(a^4-2a)^3+(a^3+1)^3+(2a^3-1)^3=(a^4+a)^3$

There are more at http://sites.google.com/site/tpiezas/010 Section 2.

This is Number Theory, not real analysis.

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How does that solve the problem? Ie, how do you get $a$ (not $a^3$) on the RHS? –  lhf May 5 '11 at 13:16
    
@Gerry: Ok, one can always edit the Tags if he/she feels that it's wrong. –  user9413 May 5 '11 at 13:47
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@lhf: In the page linked by Gerry you have several solutions and links as: $(m^3-3^6n^9)^3 + (-m^3+3^5mn^6+3^6n^9)^3 + (3^3m^2n^3+3^5mn^6)^3$ $= m(3^2m^2n^2 +3^4mn^5+3^6n^8)^3$. –  Esteban Crespi May 5 '11 at 13:58
    
@Esteban, ah, that's more like it! Perhaps you should promote this comment to an answer. –  lhf May 5 '11 at 14:01
    
Apologies to all, the equation I posted is relevant to $f^3+g^3+h^3=1$, and not to what was actually wanted, $f^3+g^3+h^3=x$ –  Gerry Myerson May 6 '11 at 0:42
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You may be interested to know identities are known that can bring this higher:

$a_1(x)^3+a_2(x)^3+a_3(x)^3 = x$

$a_1(x)^5+a_2(x)^5+a_3(x)^5+...+a_6(x)^5 = x$

$a_1(x)^7+a_2(x)^7+a_3(x)^7+a_4(x)^7+...+a_8(x)^7 = x$

where the given number of addends $a_i$ are rational functions in terms of x, hence can be treated as "Waring-like problems". The case k = 3 mentioned by Esteban is cited in Yuri Manin's book Cubic Forms (but has been solved earlier by Ryley), while k = 5 and 7 have been solved by Choudhry. Interestingly, k = 7 involves polynomials where the coefficients have 33 digits! One has to solve the simultaneous equations,

$a^2+m^5b^2 = c^2+m^5d^2$

$a^4+m^3b^4 = c^4+m^3d^4$

which, for m = 2, can be reduced to an elliptic curve and has an infinite number of integer solutions (after multiplying out the denominators), though the "smallest" one has 33 digits.

P.S. If anybody can solve k = 9, I'll be interested to know.

See, "Waring-Like Problems" at http://sites.google.com/site/tpiezas/001b.

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Here is a start:

As ihf observed, you need

$f(x)^3+g(x)^3+h(x)^3=x q(x)^3$

On the left side you have three polinomials of degrees multiple of 3, on the right the degree is M3+1, thus something must cancel on the left side.

Without loss of generality, you can consider 2 cases:

Case 1: $deg(f)=deg(g) = deg(h)$. In this case, let $ax^k, bx^k$ and $cx^k$ be the dominant monomials in $f,g,h$. Then

$$a^n+b^n+c^n=0 \,.$$ which is not possible by Fermat last Theorem. So this is not possible.

Case 2: $deg(f)=deg(g) \geq deg(h)+1$.

Then $f=ax^k+bx^{k-1}+...$. In order for the first two terms to cancel you need $g=-ax^k-bx^{k-1}+..$. And I am stucked here.

This suggest that the simplest potential example would be something like:

$$f=x^2+ax+b \,;\, g=-x^2-ax+c \,;\, h(x)=dx+e \,.$$

The computations seem too long :)

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To the question added later:

$f(x)^3 + g(x)^3 = x$

It is not possible.

Setting $x$ to be $1^3, 2^3, \dots $ gives us a non-trivial solution to the diophantine equation $x^3 + y^3 = z^3$, since $f$ and $g$ can only have finitely many roots.

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i am not able to comprehend your solution. –  user9413 May 13 '11 at 10:21
    
Are you using FLT –  user9413 May 13 '11 at 10:26
    
@Chandru: Yes, using Fermat's last theorem. –  Aryabhata May 13 '11 at 11:16
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