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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.


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).

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  • $\begingroup$ What about $a=-b$ and $c=-d$? These seem trivial, but they are not excluded by your criteria $\endgroup$ – gd1035 Jul 16 '18 at 20:49
  • $\begingroup$ Sorry, yes, I should exclude that too. $\endgroup$ – wonderich Jul 16 '18 at 20:50
  • $\begingroup$ I edited it - many thanks @gd1035 $\endgroup$ – wonderich Jul 16 '18 at 20:52
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    $\begingroup$ Intuitively, it seems to me that the $(a,b)\neq(c,d)$ condition cannot be met, but I have not yet proved it to myself satisfactorily. Here is my thinking. You have correctly deduced that $ab=cd$. Thus, the numbers $ab$ and $cd$ have the same prime factorization. It is unclear to me that you can segregate the members of the string of prime factors in two distinct ways and have the sums of those two be equal, i.e. $(a+b)=(c+d)$. If I arrive at a proof, I will post it. $\endgroup$ – Keith Backman Jul 17 '18 at 1:04
  • $\begingroup$ @Keith Backman, thanks for the comment +1 $\endgroup$ – wonderich Jul 17 '18 at 1:31
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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).

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  • $\begingroup$ thanks +1, this is a great note to notice no nontrivial solutions for this case. $\endgroup$ – wonderich Jul 17 '18 at 21:02
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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 $d$.

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  • $\begingroup$ This is true, +1, but this does not give any information about the solutions. This is tautology without any hint to the refusal or existence of nontrivial solutions. $\endgroup$ – wonderich Jul 17 '18 at 0:23
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    $\begingroup$ The answer shows that the sets $\{a,b\}$ and $\{c,d\}$ are the same as they both are the zeroes of the same quadratic polynomial. $\endgroup$ – Jens Schwaiger Jul 17 '18 at 2:52
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    $\begingroup$ @wonderich This IS the solution. I tried to spell it out in my answer. Please comment if you don't understand. $\endgroup$ – Jyrki Lahtonen Jul 17 '18 at 5:15
  • $\begingroup$ I see - thanks. this is a good note. I missed that when I read first time $\endgroup$ – wonderich Jul 17 '18 at 21:05
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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 $a=d,b=c$.


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).

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  • $\begingroup$ thanks +1, this is a great note to notice no nontrivial solutions for this case. $\endgroup$ – wonderich Jul 17 '18 at 21:03
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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?

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  • $\begingroup$ Your solution is fine, but it gives $k=a-b$, thus $$ c=2a-b=-d, $$ which is kind of a trivial answer, since $c=-d.$ $\endgroup$ – wonderich Jul 16 '18 at 21:33
  • $\begingroup$ @wonderich That is in fact the point. :-) (Essentially all of the solutions are trivial; this is just another good way of seeing that.) $\endgroup$ – Steven Stadnicki Jul 16 '18 at 21:35
  • $\begingroup$ Your solution is fine, but there is another way to say, it gives $k=b-a$, thus $$ c=b,d=a, $$ which is kind of a trivial answer. $\endgroup$ – wonderich Jul 16 '18 at 21:35
  • $\begingroup$ but thanks for the comment +1 $\endgroup$ – wonderich Jul 17 '18 at 1:32
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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).

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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

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  • $\begingroup$ Thanks! Are you sure these generate nontrivial integer solutions we demanded at first? And what are examples? $\endgroup$ – wonderich Jul 16 '18 at 20:54
  • $\begingroup$ Hmmm not sure I don't have a way to easily check these $\endgroup$ – Henry Lee Jul 16 '18 at 20:55
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    $\begingroup$ but thanks for the comment +1 $\endgroup$ – wonderich Jul 17 '18 at 1:32

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