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I came across this problem today. I would be interested to see if anyone knows a proof for it:

If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.

J. M. ain't a mathematician
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icobes
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    Are $a,b,c,d$ integers? (For instance choosing $a=3,b=4$ and $c=5$ gives $d = - \sqrt[3]{34}$. And clearly $a+b \neq c + d$). –  Nov 23 '11 at 06:07
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    Are you sure the second condition is not $a^2 + b^2 = c^2 \color{Red}{+ d^2}$? – Srivatsan Nov 23 '11 at 06:08
  • It is. It must have gotten lost in the edit. – icobes Nov 23 '11 at 06:09
  • Maybe $a^2+b^2 = c^2$ should be $a^2+b^2 = c^2+d^2$? – Paul Nov 23 '11 at 06:09
  • Under the hypotheses, you may be able to reduce to the case where $a = 0$, and hence you need to show that $b^3 = c^3 + d^3$ and $b^2 = c^2 + d^2$ implies that $b = c+d$. However, the solutions to $b^3 = c^3 + d^3$ are the trivial ones (this is a special case of Fermat's last theorem). This gives $b = c + d$. – JavaMan Nov 23 '11 at 21:45
  • @JavaMan, you seem to be assuming the variables are integers, which assumption may not be justified. – Gerry Myerson Nov 23 '11 at 22:05
  • @GerryMyerson: That is true. I meant to assume that a,b,c,d are integers. The problem seems too easy without the condition, yet too difficult with it. – JavaMan Nov 23 '11 at 22:19

3 Answers3

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If negative values of the variables are allowed then there are counterexamples. The simplest is the following:

$$a=2,\quad b=2,\quad c=\sqrt{3}-1-\sqrt{2\sqrt{3}}\doteq -1.129,\quad d=\sqrt{3}-1+\sqrt{2\sqrt{3}}\doteq2.593$$

with $a+b=4$, $c+d\doteq1.464\ .$

In the following we assume $0\leq a\leq b$ and $0\leq c\leq d$. Then icobes' conjecture is true:

Since $a^2+b^2=c^2+d^2=:r^2>0$ we may write

$$a=r\sin\bigl({\pi\over4}-\alpha\bigr), \quad b=r\sin\bigl({\pi\over4}+\alpha\bigr), \quad c=r\sin\bigl({\pi\over4}-\beta\bigr), \quad d=r\sin\bigl({\pi\over4}+\beta\bigr)$$

for some $\alpha, \beta\in\ \bigl[0,{\pi\over4}\bigr]$. It follows that $$a+b=\sqrt{2}r\cos\alpha,\quad c+d=\sqrt{2}r\cos\beta$$

and therefore

$$2(a^3+b^3)=3(a+b)(a^2+b^2)-(a+b)^3=\sqrt{2}r^3(3\cos\alpha-2\cos^3\alpha)=:\sqrt{2}r^3 f(\alpha)\ ;$$

and similarly $2(c^3+d^3)=\sqrt{2}r^3f(\beta)$.

Now $f'(t)=3\sin t\cos(2t)>0$ for $0<t<{\pi\over 4}$, whence $f$ is strictly increasing on $\bigl[0,{\pi\over4}\bigr]$. It follows that $a^3+b^3=c^3+d^3$ implies $\alpha=\beta$, and this in turn implies $a=c$, $b=d$.

Christian Blatter
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It is false.

Take $(c,d)=(1,1)$. Then $a^2 + b^2 = a^3 + b^3 = 2$ has real solutions with $(a+b) \neq 2$.

zyx
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The statement has a natural geometric interpretation when $a, b, c, d \geq 0$. A circle $x^2 + y^2 = r$ and a curve $x^3 + y^3 = s$ for $x, y > 0$ are symmetric about the line $x = y$ and if they intersect, they either intersect at a point $(x,x)$ or two symmetric points $(x,y)$ and $(y,x)$ for $x \neq y$. Thus if $a^2 + b^2 = c^2 + d^2$ and $a^3 + b^3 = c^3 + d^3$ for $a, b, c, d \geq 0$, then $(a,b) = (c,d)$ or $(a,b) = (d,c)$. In either case, $a + b = c + d$.

Why the curves intersect the way they do: If $p > q$ then $x^p + y^p = 1$ is "fatter" than $x^q + y^q = 1$, and so scaled versions of these curves will intersect as above. Maybe to prove this might require algebra like in Sivaram's answer, but the picture seems clear (to me anyway).

Zarrax
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