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