Obviously, $c\geq 6$, because otherwise $3^a+2^b$ would be negative, which is not valid. If $a=0$, then $2^b+ 2016 \equiv 3c!\equiv 0 \pmod 3\implies 2^b\equiv 0 \pmod 3$, which is not possible. If $b=0$, then $3^a+ 2016 \equiv 3c!\equiv 0 \pmod 6\implies 3^a\equiv 0 \pmod 6$, which is not possible. Hence, $a, b\geq 1$ and $c\geq 6$. Mod $3$ implies that $(-1)^b\equiv 1 \pmod 3$, which means that $b$ is even. Mod $4$ implies that $(-1)^a\equiv 1 \pmod 4$, which means that $a$ is even. Hence, we can use mod $9$ and we obtain that $2^b\equiv 1 \pmod 9$, which means that $b\equiv 0 \pmod 6$, let $b=6k$ and $a=2m$. Now if $c\geq 7$, mod $7$ gives us that $3^a+64^k+2015\equiv 3^a+1+(-1) \equiv 0 \pmod{7} $, hence no solutions, if $c\geq 7$. Case when $c=6$: gives that $9^m+2^{6k}+2015=3\cdot 6! \Longleftrightarrow 9^m+2^{6k}=145$. If $k\geq 2$, then $LHS>2^{12}>1000>145=RHS$, no solutions. If $k=1$, then $9^m=81\implies m=2$, hence $(a,b,c)=(4,6,6)$ is a solution. Answer. $\boxed{(a,b,c)=(4,6,6)}$.

rafaello wrote: Obviously, $c\geq 6$, because otherwise $3^a+2^b$ would be negative, which is not valid. If $a=0$, then $2^b+ 2016 \equiv 3c!\equiv 0 \pmod 3\implies 2^b\equiv 0 \pmod 3$, which is not possible. If $b=0$, then $3^a+ 2016 \equiv 3c!\equiv 0 \pmod 6\implies 3^a\equiv 0 \pmod 6$, which is not possible. Hence, $a, b\geq 1$ and $c\geq 6$. Mod $3$ implies that $(-1)^b\equiv 1 \pmod 3$, which means that $b$ is even. Mod $4$ implies that $(-1)^a\equiv 1 \pmod 4$, which means that $a$ is even. Hence, we can use mod 9 and we obtain that $2^b\equiv 1 \pmod 9$, which means that $b\equiv 0 \pmod 8$, let $b=8l$ and $a=2m$. Now mod $10$ gives us that $3^a+2^b+2015\equiv (-1)^m+256^l+5\equiv (-1)^m +1\equiv 0 \pmod{10} $, hence $m$ is odd. Mod $16$ gives us that $3^a+2^b+2015\equiv 0 \pmod{16} $, hence $9^m+15\equiv 8\not\equiv 0 \pmod{16} $, since $m$ is odd. Thus, there are no solutions. $(a,b,c)=(4,6,6)$ is a solution?

mathleticguyyy wrote: rafaello wrote: Obviously, $c\geq 6$, because otherwise $3^a+2^b$ would be negative, which is not valid. If $a=0$, then $2^b+ 2016 \equiv 3c!\equiv 0 \pmod 3\implies 2^b\equiv 0 \pmod 3$, which is not possible. If $b=0$, then $3^a+ 2016 \equiv 3c!\equiv 0 \pmod 6\implies 3^a\equiv 0 \pmod 6$, which is not possible. Hence, $a, b\geq 1$ and $c\geq 6$. Mod $3$ implies that $(-1)^b\equiv 1 \pmod 3$, which means that $b$ is even. Mod $4$ implies that $(-1)^a\equiv 1 \pmod 4$, which means that $a$ is even. Hence, we can use mod 9 and we obtain that $2^b\equiv 1 \pmod 9$, which means that $b\equiv 0 \pmod 8$, let $b=8l$ and $a=2m$. Now mod $10$ gives us that $3^a+2^b+2015\equiv (-1)^m+256^l+5\equiv (-1)^m +1\equiv 0 \pmod{10} $, hence $m$ is odd. Mod $16$ gives us that $3^a+2^b+2015\equiv 0 \pmod{16} $, hence $9^m+15\equiv 8\not\equiv 0 \pmod{16} $, since $m$ is odd. Thus, there are no solutions. $(a,b,c)=(4,6,6)$ is a solution? Oh yes, my bad, it is a solution. @above, now fixed.

The only solution is $(a,b,c)=(4,6,6)$ Assume that $c \geq 7$. Then we must have that $3^a+2^b \equiv 1 \pmod{7}$. Also we must have that $3^a+2^b \equiv 1 \pmod{8}$. By examining the two conditions we must have that $a=6k+4$ and that $b=3t+2$. Since $a \geq 4$, we must have that $2^{3t+2} \equiv 1 \pmod{9}$. But notice that $2^{3t+2} \equiv \{ 4,5\} \pmod{9}$. Thus we get a clear contradiction on the assumption that $c \geq 7$. Thus we must have that $c \leq 6$. Now assume that $c \leq 5$, then we must have that $LHS \leq 3c! \leq 3.120 = 360$, but this implies that $3^a+2^b$ must be negative, which doesn't hold. Thus we must have that $3^a+2^b =145$, which gives us that $a=4$ and $b=6$.

After seeing that $a,b$ are even, checking for $c \geq 31$, $modulo 31$ gives a contradiction, and we can then check for $c=6,7, ... , 30$, It is somewhat <<lazy>> , but it works.

If $c\le5$, $LHS>RHS$ thus $c\ge6$, so from now on when we cite $c$ we have that $c\ge6$ Inspecting $\pmod 3$ yields $3^a+2^b+2015\equiv (-1)^b+2\pmod 3$ thus $b$ is even. Furthermore if we take $\pmod 4$ we obtain $3^a+2^b+2015\equiv (-1)^a+3\pmod 4$ thus $a$ is also even. So, now let $a=2n\text{ and }b=2k$ Thus the equation transforms into $9^n+4^k+2015=3c!$ Furthermore taking $\pmod 5$ we obtain $9^n+4^k+2015\equiv (-1)^n+(-1)^k\pmod 5$ which implies that $n$ and $k$ are pairwise distinct. Inspecting $\pmod 9$ yields $9^n+4^k+2015\equiv -1,1\text{ or }4-1\pmod 9\Longrightarrow 4^k\equiv 1\pmod 9$ thus $k$ is a multiple of $3$ and let $k=3x$ Moreover the equation transforms into $9^n+64^x+2015=3c!$ Also notice that if we take $\pmod 7$ we obtain $9^n+64^x+2015\equiv2^n+1-1\equiv 2^n\equiv 1,2\text{ or }3\pmod 7$, however $3c!\equiv 0\pmod 7$ for $c\ge7$, thus $c\le6$ which implies that $c=6$, plugging into our equation we obtain $9^n+64^x+2015=2160\Longrightarrow 9^n+64^x=145$ Furthermore notice that if $x\ge2,$ $LHS>RHS$, thus $x=1$ which implies that $9^n=81\Longrightarrow n=2$. Plugging into our equations for $a$ and $b$ we obtain $a=4\text{ and }b=6$, thus $(a,b,c)=(4,6,6)$ is the only solution. In conclusion $\boxed{(a,b,c)=(4,6,6)\text{ is the only solution in nonnegative integers}}$ $\blacksquare$.