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I was challenged to prove that there are infinitely many solutions to the equation$$a^3+b^3+c^3+d^3=3\ \ \text{ with }(a,b,c,d)\in\mathbb Z^4$$

That was easy: elementary algebra is enough to prove that $\forall z\in \mathbb Z$ then $$(a,b,c,d)=(1+6z^3,\,1-6z^3,\,-6z^2,\,1)$$ is a solution. Call these solutions and their permutations trivial.

Define a nonsum as a value of $k\in\mathbb Z$ for which there is no solution with $a+b=k$.

What about proving (or disproving/perfecting) the following conjectures?

  1. If polynomials $A$, $B$, $C$, $D$ with integer coefficients are such that $\forall z\in \mathbb Z,\ (A(z),B(z),C(z),D(z))$ is a solution, and one such solution is non-trivial, then all the polynomials are constant.
  2. There are nevertheless infinitely many non-trivial solutions.
  3. Any $k\equiv0\pmod3$ is a nonsum.
  4. $k=7$ is a nonsum.
  5. There are infinitely many nonsums with $k\not\equiv0\pmod3$
  6. If $k$ is a nonsum, then $-k$ is a nonsum.
  7. If $k$ is not a nonsum (that is, if there exists a solution with $a+b=k$), then there are infinitely many solutions with $a+b=k$.
Tito Piezas III
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fgrieu
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3 Answers3

4

Conjecture 2 is true.

Given an initial solution to $$a^3+b^3+c^3+d^3=N$$

one can generate infinitely more using a Pell equation. Start with the general identity, $$t^3+x^3+y^3+z^3 = (a^3+b^3+c^3+d^3)\big(p^2+wq^2\big)^3$$

where, $$\begin{aligned} t &= a p^2 - (c^2-d^2) p q + b w q^2 \\ x &= b p^2 + (c^2-d^2) p q + a w q^2\\ y &= c p^2 + (a^2 - b^2) p q + d w q^2\\ z &= d p^2 - (a^2 - b^2) p q + c w q^2\\ w &= (a+b)(c+d) \end{aligned}$$

with $(a,b,c,d)$ suitably permuted so that $w$ is negative, then $p,q$ chosen to solve the Pell equation, $$p^2+wq^2 = 1$$

Since from your parameterization, $$(a,b,c,d)=(1+6z^3,\,1-6z^3,\,-6z^2,\,1)$$ we get a negative value for $w$, $$w=(a+b)(c+d) = -2(6z^2-1)$$ necessary for the Pell equation, this shows we can derive infinitely many non-trivial solutions from the "trivial" one. Perhaps you can use this on some of your other conjectures.

Tito Piezas III
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    Incidentally, for the special case $a^3+b^3+c^3+d^3 = 0$, then $p,q$ become free parameters, yielding infinitely many quadratic parameterizations to the taxicab problem. – Tito Piezas III Feb 12 '18 at 12:35
2

Conjecture 3 is true.

Note first that for any integer $n$, $n^3 \equiv -1,0$, or $1\pmod 9$. Hence any solution of $a^3+b^3+c^3+d^3=3$ must contain three cubes congruent to $1\pmod 9$ and one cube congruent to $0\pmod9$.

Now suppose $a+b = k\equiv0\pmod3$. We have:

$$a^3+b^3=(a+b)^3-3ab(a+b)$$

where:

$$(a+b)^3\equiv0\pmod{3^3}\quad \text{and}\ \textit{a fortiori}\quad \equiv0\pmod9$$

$$3ab(a+b)\equiv0\pmod9$$

Hence $a^3+b^3\equiv0\pmod9$ which is inconsistent with the condition stated above, since it implies either both $a^3$ and $b^3$ congruent to $0\pmod9$ or one of them congruent to $-1$ and the other to $1\pmod9$.

Similar but not identical reasoning shows that conjecture 4 is true, and more generally any $k\equiv0\pmod7$ is a nonsum.

For any integer $n$, $n^3 \equiv -1,0$, or $1\pmod 7$. Hence any solution of $a^3+b^3+c^3+d^3=3$ must contain either (Case 1) three cubes congruent to $1\pmod 7$ and one cube congruent to $0\pmod7$ or (Case 2) four cubes congruent to $-1\pmod7$

Now suppose $a+b = k\equiv0\pmod7$. We have:

$$a^3+b^3=(a+b)(a^2-ab+b^2)\equiv0\pmod7$$

But this is inconsistent with both Case 1 and Case 2 above, since it implies either both $a^3$ and $b^3$ congruent to $0\pmod7$ or one of them congruent to $-1$ and the other to $1\pmod7$.

Adam Bailey
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0

Equation given above is shown below; $a^3+b^3+c^3+d^3=3$

In the solution given by Tito Piezas for $z=1$ we have $(a,b,c,d)=(7,-5,-6,1)$

and $(p,q,w)=(19,6,-10)$ which in turn gives $(t,x,y,z)=(337,-335,210,-215)$

Thus one of the many new solutions is,

$(t^3+x^3+y^3+z^3) = (a^3+b^3+c^3+d^3)(p^2+wq^2)^3=3$

Sam
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