Proving that for integers $a \ge 5, a > b, b \ge 1$, $2^a - 27 \nmid 2^b + 15$ I am trying to show that for integers $a \ge 5, a > b, b \ge 1$:
$$2^a - 27 \nmid (2^b + 15)$$
Is the following inductive argument valid?
Please let me know if I made any mistake or if there is a more standard way to prove the same point.
(1)  Base Case: $a=5$:
$32 - 27 = 5 \nmid (2^b + 15)$ since $2^b \not\equiv 0 \pmod 5$
(2)  Assume that the assumption is true for any $a \ge 5$
(3)  Inductive Case:

*

*Assume that $2^{a+1} - 27 | (2^b + 15)$ with $a+1 > b$


*$2^{a+1} - 42$ is not a power of $2$

*

*Assume that $2^{a+1} - 42 = 2^c$

*$2^c(2^{a+1-c} - 1) = 42$

*Since $2^c | 42$, $c=1$

*$2^{a+1-c} = 22$ which is impossible since $11 \nmid 2^{a+1-c}$

*We can reject the original assumption.



*$2^b \equiv 2^{a+1}-42 \pmod {2^{a+1}-27}$


*$2^b > 2^{a+1} - 27$  Since $2^{a+1}-42$ is not a power of $2$


*Since $2^a > 27$, it follows that $2^b > 2^{a}$  Since $2^a < 2^a + (2^a - 27)$


*But then $b > a$ which contradicts the assumption that $b < a+1$


*So we reject the assumption that $2^{a+1} - 27 | (2^b + 15)$

Your proof appears to be basically correct. Nonetheless, a minor point is your inductive case doesn't actually technically involve induction as there's no actual use of the inductive hypothesis, i.e., that $2^a - 27 \not\mid 2^b + 15$, so induction was not required.

Alternatively, using size comparisons similar to what you did, here is what I consider to be a somewhat simpler & more direct proof. First, note that
$$2^a - 27 \mid 2^b + 15 \implies 2^a - 27 \le 2^b + 15 \implies 2^a \le 2^b + 42 \tag{1}\label{eq1A}$$
Next, with $a \gt b \implies a \ge b + 1 \implies 2^a \ge 2^{b+1} = 2(2^b)$, then \eqref{eq1A} becomes
$$\begin{equation}\begin{aligned}
2(2^{b}) & \le 2^{a} \le 2^b + 42 \\
2^{b} & \le 42 \\
2^{b} + 42 & \le 84
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Using this with the right side inequality in \eqref{eq1A} gives
$$2^a \le 84 \implies a \le 6 \tag{3}\label{eq3A}$$
You've already shown $a = 5$ doesn't work. With $a = 6$, then $2^a - 27 = 64 - 27 = 37$. However, $2^b + 15$ is not a multiple of $37$ for any $b \lt 6$. This therefore shows your conjecture is always true.
