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?
 A: Write $a = 2^x \cdot m$ and $b = 2^y \cdot n$ where $m$ and $n$ are odd. WLOG $x \geq y$. Now,
$$2^{3x} \cdot m^3 + 2^{3y} \cdot n^3 = 2^{3y} \cdot (2^{3x-3y} \cdot m^3 + n^3) = 2^c$$ If $x > y$, we have a contradiction since in that case an odd number divides LHS. So, $x = y$ and so you would arrive at your argument.
A: If $a$ and $b$ are both even, say $a=2^mA$ and $b=2^nB$ with $A$ and $B$ odd, then without loss of generality $n\geq m$ and so
$$2^c=a^3+b^3=2^{3m}\big(A^3+2^{3(n-m)}B^3\big),$$
from which it follows that
$$2^{c-3m}=A^3+2^{3(n-m)}B^3.$$
Because $A$ and $B$ are positive we see that $2^{c-3m}>1$, and so the left hand side is even. Because $A$ is odd it follows that also $2^{3(n-m)}B^3$ is odd, and so $n=m$. Then for $C=c-3m$ we have
$$A^3+B^3=2^C,$$
thus yielding a new solution with $A$ and $B$ both odd.
If $a$ and $b$ are both odd then you already observed that the factorization
$$2^c=a^3+b^3=(a+b)(a^2-ab+b^2),$$
implies that $a^2-ab+b^2=1$. Viewing this as a quadratic in $a$ shows that
$$a=\frac12\big(b\pm\sqrt{b^2-4(b^2-1)}\big)=\frac12\big(b\pm\sqrt{4-3b^2}\big).$$
For this to be an integer we must have $4-3b^2>0$, so $b=1$ as $b$ is positive by assumption. Then also $a=1$ as $a$ is positive by assumption. This shows that $a=b=1$. It follows that every solution is of the form $a=b=2^m$ for some nonegative integer $m$.
A: 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.
A: 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 $a,b = 1$.
