<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] It's easy to see if $a= 2^km; m$ odd and $b = 2^jn;n $odd that $k=j$ (we can factor out $2^{\min(3j,3k)}$ from $a^3 + b^3$) so wolog we can assume $a,b$ are both odd.
If $a \ne b$ and wolog $a < b$ we have $a= 2^x -d;b =2^x + d$ where $d$ is odd and so
$(a^2 -ab+b^2) = (2^x-d)(2^x-d) - (2^x+d)(x^2 - d) +(2^x+d)(2^x+d)=$
$4\cdot 2^{2x} +3d^2$ is .... odd.
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