Integers in the form $\pm 2^a \pm 3^b \pm 5^c$

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 $$\alpha,\beta,\gamma \in \{-1,1\}$$?

• Do you have a reason to believe this is the case? – Carl Schildkraut Apr 4 at 20:49
• 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. – Eevee Trainer Apr 4 at 20:51
• Definitively not: it is only curiosity. I just wanted to know whether there exist methods to attack suck kind of questions. – Paolo Leonetti Apr 4 at 20:51
• When $a\ne 0$, the sum is always even. – FredH Apr 4 at 21:04
• @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 $\beta,\gamma<0$). – Paolo Leonetti Apr 4 at 21:36

In particular, if we are thinking mod $$120$$, then
$$\pm2^a \in \{ 1, 2, 4, 8, 16, 32, 56, 64, 88, 104, 112, 116, 118, 119 \}$$ $$\pm3^b \in \{ 1, 3, 9, 27, 39, 81, 93, 111, 117, 119 \}$$ $$\pm5^c \in \{ 1, 5, 25, 95, 115, 119 \}$$
$$\{19, 47, 49, 59, 61, 71, 73, 101\}$$
• so, despite the fact that I guessed $120$ basically at random, the only smaller modulus that is a counterexample is $90$. – Gregory Nisbet Apr 4 at 21:27