If $a,b$ and $c$ are positive integers such that $2^a+2^b+2^c=148$ then solve for $a,b,c$.
My Attempt
Clearly any permutation of $<2,4,7>$ is a solution.
$2^a+2^b=148-2^c\geq 4\Rightarrow 2^c\leq 144\Rightarrow c\leq 7$
But how do we prove that the only solution is permutation of $<2,4,7>$ without resorting to Hit and Trial
Quick argument:
$$148=10010100_2$$ which gives rise to the expansion you noted.
Base $2$ representations are unique, so this is the only triple with distinct terms.
Thus we need only consider cases in which two or more of $(a,b,c)$ coincide.
If all three coincided, $3$ would divide $148$ so that's impossible.
If two coincided, then we remark that $2\times 2^n=2^{n+1}$ so we'd have a shorter expansion in powers of $2$. If the third exponent was $n+1$, we'd have $$148=2^n+2^n+2^{n+1}=2^{n+2}$$ which isn't possible. If the third exponent was not $n+1$ then we'd get a binary expansion for $148$ with only two $1's$ and, again, that would contradict the uniqueness of binary expansions.
Let $a$ be the smallest of the three. Then we can write: $$2^a\cdot (1+2^{b-a}+2^{c-a})=2^2\cdot 37.$$
If say $a=b$ then we get that $37 \mid 2+2^{c-a}$ and it’s easy to check that $37-2,$ $74-2,$ $111-2$ or $148-2$ are not powers of $2.$ So $1+2^{b-a}+2^{c-a}$ is an odd integer and we deduce that $$a=2,$$ $$1+2^{b-a}+2^{c-a}=37.$$
Or we can write $$2^{b-a}+2^{c-a}=36.$$
Let $b-a \leq c-a.$ Then we can do the same as in the first step: $$2^{b-a}(1+2^{c-b})=2^2\cdot 9.$$
$1+2^{c-b}$ must be odd so we get that $b-a=2$ and therefore $b=4.$ Also $$1+2^{c-b}=9.$$ Now we easily find $c=3+b=7.$ So yes, $\{2,4,7\}$ is the only set of numbers.
