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I was browsing through facebook and came across this image: enter image description here

I was wondering if we can find more examples where this happens?

I guess this reduces to finding integer solutions for the equation

$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for integers a,b,c

Or can we even further extend to when they are all distinct that is finding solutions to

$$ \frac{a^3+b^3}{c^3+d^3} = \frac{a+b}{c+d} $$ for integers a,b,c

I don't really have that much knowledge in the number theory area so I have come here

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  • $\begingroup$ It holds when $b+c = a$, assuming $a, b, c$ are distinct natural numbers. More examples. $\frac{3^3+1^3}{3^3+2^3} = \frac{3+1}{3+2}$, $\frac{7^3+2^3}{7^3+5^3} = \frac{7+2}{7+5}$ $\endgroup$ – Jakobian Aug 5 '18 at 9:59
  • $\begingroup$ $$\frac{a^3+b^3}{c^2+d^3}=\frac{(a+b)(a^2-ab+b^2)}{(c+d)(c^2-cd+d^2)}=\frac{a+b}{c+d}$$ $$a^2-ab+b^2=c^2-cd+d^2$$ $\endgroup$ – individ Aug 5 '18 at 9:59
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    $\begingroup$ artofproblemsolving.com/community/c3046h1048192___ $\endgroup$ – individ Aug 5 '18 at 10:08
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$$\frac{a^3+b^3}{a^3+c^3}=\frac{(a+b)(a^2-ab+b^2)}{(a+c)(a^2-ac+c^2)}=\frac{a+b}{a+c}$$

If $a+c \neq 0$ and $a+b \neq 0,$ then $$a^2-ab+b^2=a^2-ac+c^2,$$namely $$(b+c-a)(b-c)=0.$$

If $b=c$, the case is trivial. If $b \neq c$, then $$b+c=a.$$

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Let $S$ be the set of allowed values of $a$, $b$, and $c$ ($S=\mathbb{Z}$ in this OP's setting, but $S$ can be something else like $\mathbb{Q}$, $\mathbb{Q}_{>0}$, $\mathbb{R}$, or even $\mathbb{F}_p$, where $p$ is a prime natural number). If $b=-a$, then $c$ can be any number not equal to $-a$. That is, $(a,b,c)=(a,-a,c)$ with $c\neq -a$ is always a solution. From now on, we assume that $b\neq-a$.

Then, as invidid found, $$\frac{a^2-ab+b^2}{a^2-ac+c^2}=\left(\frac{a^3+b^3}{a^3+c^3}\right)\left(\frac{a+b}{a+c}\right)^{-1}=1\,.$$ Hence, $a^2-ab+b^2=a^2-ac+c^2$, or $$(b-c)(b+c-a)=0\,.$$ That is, $b=c$ or $a=b+c$.

This concludes that all solutions $(a,b,c)\in S^3$ takes the form $(a,-a,c)$ with $c\neq -a$, $(a,b,b)$ with $b\neq -a$, and $(b+c,b,c)$ with $c\neq -\frac{b}{2}$. You can check that these indeed are solutions. The two solutions $(a,b,c)=(5,2,3)$ and $(a,b,c)=(579,123, 456)$ that you found are of the form $(a,b,c)=(b+c,b,c)$.

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