2023/032) Given positive integers a,b,c and $a^3+b^3 = 2^n$ prove that $a=b$

Both a nd b have to be even or add as RHS is even as it has to be > 1.

Because if a is odd and b even or vice versa then LHS is odd.

Now let us consider 1st case that is a and bare odd

So we get

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

a+b is even and $a^2-ab+b^2= a(a-b) + b^2$ is odd and it has to be 1

Or $a(a-b) + b^2=1$ 

The above is possble only if a-b = 0(else  or a = b and we get a = b = 1

Now of the even case 

Let $a=p.2^q$ and b= $m.2^r$ and without loss of generality let$ q >= r$ and p and m are odd.

So we get $p^32^{3q} + m^32^{3r} = 2^n$

Devideing both sides by $2^{3r}$ we get

$p^32^{3q-3r} + m^3 = 2^{n-3r}$

RHS is even as it is minimum 2

LHS is even if 3q=3r or q = r

Puttting it we get $p^3+q^3 = 2^{n-3r}$ 

As p and q are odd so both are same and hence $a= b = 1$ for odd and $a=b=2^x$ for even