<s>[INST] Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$. 
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.

I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?
 [/INST] HINT:

*

*For the case where $a,b$ both even, factor out the largest power of $2$ from both of them. You will either reduce to exactly one of $a'$, $b'$ odd [which you already noted is impossible] or both $a,b$ odd. To elaborate a bit more, let $j$ be the largest integer s.t. $2^j$ divides both $a$ and $b$. Then write $2^ja'=a$ and $2^jb'=b$, with at least one of $a',b'$ odd.
Then $$(a')^3+(b')^3 = 2^{3j}2^{c-3j},$$ so as at least one of $a',b'$ is odd, then by your reasoning above, both $a'$ and $b'$ must be odd.


*For the case where $a,b$ both odd: You noted that this implies the equation
$$(a+b)(a^2-ab+b^2)= 2^c.$$ So then both $(a^2-ab+b^2)$ and $(a+b)$ must be powers of $2$. However, $a^2-ab+b^2$ is odd [why is this] so this gives $a^2-ab+b^2=1$, because $1$ is the only odd power of $2$. So this gives $(a+b)=(a^3+b^3)$ [why is this] which
is only possible [assuming $a,b$ positive integers] for $a,b = 1$.
</s>