Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$ and $a+b=c+d$

Question

The goal is to understand a set of nontrivial solutions of

$$ a^3+b^3=c^3+d^3 \neq 0, \tag{1} $$ $$ a+b=c+d \neq 0, \tag{2} $$ for $a,b,c,d\in \mathbb{Z}$ where we demand $(a,b)\neq (c,d)$ and $(a,b)\neq (d,c)$. Also $a\neq -b, c \neq -d$. The (1) has partial but very nice answers here.

• Question: Are there any nontrivial solutions? (e.g. $(a,b)\neq (c,d)$ and $(a,b)\neq (d,c)$.) How many simple but nontrivial solutions are there in the smaller values of $a,b,c,d$? (The simple form the solutions are the better.) Or can one prove or disprove that there are nontrivial solutions? (Even better, if we can get a quick algorithm to generate the solutions.)

• A non-trivial example or a proof of non-existence is required to be accepted as a final answer.

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Note add:

My trials/attempts: Since it is encouraged to show one's own attempt. Let me share a few comments on what these two equations boil down to:

(I) Combine (1)-(2)$^3$, with (2), we can obtain that the following is true:

$$ ab=cd, \tag{3} $$

Similarly, combining (2)$^2$-2(3), we can obtain the following is also true:

$$ a^2+b^2=c^2+d^2, \tag{4} $$

Thus, we can simply use (2) and (3) together with the constraint of $a,b,c,d\in \mathbb{Z}$ where $(a,b)\neq (c,d)$ and $(a,b)\neq (d,c)$, to get the answer.

Thank you!

p.s. When $d=0$ (or any one of the four of $(a,b,c,d)$ is zero), we know it is impossible due to the Fermat's Last Theorem. There are some nontrivial solutions for (1) here, but not enough to satisfy also (2). The naive Plato number satisfies (1) but not (2).

Answer 1

You have already deduced that if $a^3+b^3=c^3+d^3$ and $(a+b)=(c+d)=m$ then $ab=cd=n$, because $(a+b)^3-(a^3+b^3)=3ab(a+b)=(c+d)^3-(c^3+d^3)=3cd(c+d)$; dividing through by $3m$ gives that result.

Since all of the prime factors of $a$ must be found in $n$ and hence in $cd$, we can group the prime factors of $a$ as follows: Those that are to be found in $c$ multiply to $r$, and those that are to be found in $d$ multiply to $s$. (1) $a=rs$. Similarly for the prime factors in $b$; those found in $d$ multiply to $t$ and those found in $c$ multiply to $u$. (2) $b=tu$.

From this, we deduce (3) $c=ru$ and (4) $d=st$.

$(a+b)=(c+d)$ means $rs+tu=ru+st$. Rearranging, we get $r(s-u)=t(s-u)$. Plainly, $r=t$ and $rs=a=ts=d$ from which follows $b=c$. Note that if $r>1$, the terms $a,b,c,d$ have a common factor and do not constitute a primitive solution to the problem; for primitive solutions, $r=t=1$.

Pairs of integers that have a common sum and a common product are (except for order) the same. I am certainly not the first person to realize and prove this.

In sum, the condtions that $a^3+b^3=c^3+d^3$ and $(a+b)=(c+d)$ preclude the final condition that $(a,b)$ and $(c,d)$ are distinct (other than by order).

Answer 2

so we are given that: $$a^3+b^3=c^3+d^3$$ and $$a+b=c+d$$ start of with: $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$ $$\therefore (a+b)^3=(a^3+b^3)+3ab(a+b)$$ $$\therefore (c+d)^3=(c^3+d^3)+3ab(c+d)$$ then: $$3ab=\frac{(c+d)^3-(c^3+d^3)}{(c+d)}$$ Hope this helps

Answer 3

Here's another approach: $c+d=a+b$ is equivalent to saying that $c=a+k, d=b-k$ for some $k\in\mathbb{Z}$. Now, $c^3+d^3 = (a+k)^3+(b-k)^3=(a^3+b^3)+3k^2(a+b)+3k(a^2-b^2)$; this implies that $3k^2(a+b)+3k(a^2-b^2)=0$. Since $k\neq 0$ for any non-trivial solution, we can divide out by $3k$. Now, can you work the rest of the algebra?

Answer 4

According to $(2)$, let $p=a+b=c+d$, and according to $(3)$, let $q=ab=cd$. Then the two roots of $$X^2-pX+q $$ are $a$ and $b$, but also $c$ and $d$.

Answer 5

Trying to spell out the argument by others as well as adding details and links.

This is a simple case of elementary symmetric polynomials and Newton's identities.

Let's begin with any two numbers $x_1,x_2$. They are zeros of the polynomial $$ p(x)=(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2. $$ Let's denote the coefficients here $e_1=x_1+x_2$ and $e_2=x_1x_2$. Newton's identities (or Waring's formula, or just brute force expansion using the binomial theorem) imply that $$ x_1^3+x_2^3=e_1^3-3e_1e_2. $$

We have both $$a+b=c+d\qquad\text{and}\qquad a^3+b^3=c^3+d^3.\qquad(*)$$ We apply the above to the cases $\{x_1,x_2\}=\{a,b\}$ as well as $\{x_1,x_2\}=\{c,d\}$. The first equation in $(*)$ reads simply $$ e_1(a,b)=e_1(c,d) $$ but the second is slightly more complicated: $$ e_1(a,b)^3-3e_1(a,b)e_2(a,b)=e_1(c,d)^3-3e_1(c,d)e_2(c,d). $$ Anyway, the two pairs share the same value of $e_1=e_1(a,b)=e_1(c,d)$, so plugging that into the second equation gives $$ e_1^3-3e_1e_2(a,b)=e_1^3-3e_1e_2(c,d). $$ Cancelling the $e_1^3$ terms and dividing by three implies $$ e_1\,e_2(a,b)=e_1\,e_2(c,d).\qquad(**) $$ The equation $(**)$ leaves us with two cases. Either 1) $e_1=0$, or 2) $e_2(a,b)=e_2(c,d)$.

Case 1. If $e_1=0=a+b=c+d$ then $b=-a$ and $d=-c$. In this case also $a^3+b^3=0$ and $c^3+d^3=0$ so these are solutions to your system.

Case 2. If $e_2=e_2(a,b)=e_2(c,d)$, then we have the polynomial identity $$ p(x)=(x-a)(x-b)=(x-c)(x-d).$$ So $\{a,b\}$ as well as $\{c,d\}$ are the set of zeros of $p(x)$. Because a quadratic polynomial has exactly two (complex) solutions, we can conclude that $$\{a,b\}=\{c,d\}$$ as set. In other words either $a=c,b=d$, or $a=d,b=c$.

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Observe that the argument does not rely on $a,b,c,d$ being integers. They can be elements of any field (as long as the characteristic is not equal to three - we did divide by three to get the equality of $e_2$s).

Answer 6

I think that a little Galois theory will give us the shortest and simplest proof.

Let $\omega$ be a primitive 3rd root of 1. The quadratic field $\mathbf Q(\omega)$ is a cyclic extension of $\mathbf Q$, with Galois group generated by the complex conjugation $\gamma: \omega \to \omega^2$. Because of the formula $a^3+b^3=(a+b)(a+b\omega)(a+b\omega^2)$ and the analogous formula for $(c^3+d^3)$, the conditions of the problem are equivalent to $N(a+b\omega)=N(c+d\omega)$, or $N(a+b\omega /c+d\omega)=1$, where $N$ denotes the norm map of $\mathbf Q(\omega)/\mathbf Q$. Since our extension is cyclic, Hilbert's thm.90 tells us that $a+b\omega /c+d\omega$ will be of the form $\gamma(z)/z=d+e\omega^2/d+e\omega$, or equivalently, by identification, $ae=ce+df, af+be=de, bf=cf$. If $f\neq 0$, the 3rd equality reads $b=c$, and the condition $a+b=c+d$ gives $a=d$. If $f=0$, the 2nd equality reads $be=de$, hence $b=d$ because $z\neq 0$. Summarizing, the problem admits a unique solution up to a transposition of the coordinates.

NB. As already noticed by Jyrki Lahtonen, the previous argument works above any field $K$ of characteristic $\neq 3$ (this condition ensures that $K(\omega)/K$ is cyclic of degree 2).