# If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$

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 $a + b = c + d$.

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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 $a+b \neq c + d$). – user17762 Nov 23 '11 at 6:07
Are you sure the second condition is not $a^2 + b^2 = c^2 \color{Red}{+ d^2}$? – Srivatsan Nov 23 '11 at 6:08
It is. It must have gotten lost in the edit. – icobes Nov 23 '11 at 6:09
Maybe $a^2+b^2 = c^2$ should be $a^2+b^2 = c^2+d^2$? – Paul Nov 23 '11 at 6:09
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

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 $a=c$, $b=d$.

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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 $(a+b) \neq 2$.

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What's an example? – user7530 Nov 23 '11 at 10:31
If you look at wolframalpha.com/input/…, you can see that there are two other real roots. – David Speyer Nov 23 '11 at 14:48
@user7530: See my answer below. – Christian Blatter Nov 23 '11 at 21:43

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).

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