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Find all positive integer solutions to $2^a+b!=2^c$. Wolfram gives $(a,b,c)=(1,2,2),(1,3,3),(3,4,5),(3,5,7)$.

The reduction of this problem when $b=a+1$ uses techniques like $v_p(n)$, as well as Legendre's formula, $v_2(n!)=n-s_2(n)$, where $s_2(n)$ is the number of 1's in base-2 notation of n. (In fact, this reduction is a problem from Singapore Math Olympiad, with answer found here.)

Another reduction, with the condition for $a \geq b$, is easily solved.

How would this equation be solved? Thank you.

Math
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    What do you mean by "no such constraints"? – user2661923 Jun 17 '24 at 05:51
  • For what it's worth, virtually every MathSE posted question that I have seen, that followed this article on MathSE protocol has been upvoted rather than downvoted. I am not necessarily advocating this protocol. Instead, I am merely stating a fact: if you scrupulously follow the linked article, skipping/omitting nothing, you virtually guarantee a positive response. – user2661923 Jun 17 '24 at 05:51
  • @user2661923 just ignore that- I meant no additional constraints such as $a \geq b$ or $b=a+1$. – Math Jun 17 '24 at 06:29
  • İt was a nice question. But, unfortunately it is dubuqate. @MathsAddict – Bob Dobbs Jun 17 '24 at 08:38

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