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It follows by the theory of linear form in logarithms that "few" integers can be written as $2^a-3^b$, with integers $a,b\ge 0$ (see here). What about the case of more than two variables?

Question. Is it true that every sufficiently large integer can be expressed as $$ \alpha 2^a+\beta 3^b +\gamma 5^c, $$ for some integer $a,b,c \ge 0$ and $\alpha,\beta,\gamma \in \{-1,1\}$?

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    $\begingroup$ Do you have a reason to believe this is the case? $\endgroup$ – Carl Schildkraut Apr 4 at 20:49
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    $\begingroup$ Without an idea as to what constitutes "sufficiently large," I don't think it's particularly easy to say. One could hypothetically always come up with a counterexample and then, without a concrete idea of what is large enough, you could just go "think bigger." So to speak. $\endgroup$ – Eevee Trainer Apr 4 at 20:51
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    $\begingroup$ Definitively not: it is only curiosity. I just wanted to know whether there exist methods to attack suck kind of questions. $\endgroup$ – Paolo Leonetti Apr 4 at 20:51
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    $\begingroup$ When $a\ne 0$, the sum is always even. $\endgroup$ – FredH Apr 4 at 21:04
  • $\begingroup$ @FredH: this leads to another solution: odd integers are represented by force with $a=0$ and, for every choice of $\beta,\gamma$, the sums $\beta3^b+\gamma5^c$ represent just few integers (if $\beta,\gamma>0$ then the represented integers $\le x$ are $O(\log^2 x)$, and similarly if $\beta,\gamma<0$). $\endgroup$ – Paolo Leonetti Apr 4 at 21:36
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We can definitively say that there's no range of numbers above which "every number is expressible in this form" because there are certain modulus classes you can't ever hit with our sum of three numbers.

In particular, if we are thinking mod $120$, then

$$\pm2^a \in \{ 1, 2, 4, 8, 16, 32, 56, 64, 88, 104, 112, 116, 118, 119 \} $$ $$\pm3^b \in \{ 1, 3, 9, 27, 39, 81, 93, 111, 117, 119 \} $$ $$\pm5^c \in \{ 1, 5, 25, 95, 115, 119 \} $$

The unhittable classes are as follows:

$$\{19, 47, 49, 59, 61, 71, 73, 101\}$$

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  • $\begingroup$ Of course I have been unlucky with the choice of the numbers. But well, it works :) $\endgroup$ – Paolo Leonetti Apr 4 at 21:22
  • $\begingroup$ so, despite the fact that I guessed $120$ basically at random, the only smaller modulus that is a counterexample is $90$. $\endgroup$ – Gregory Nisbet Apr 4 at 21:27

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