Bariz - Problem

$\rm 5^a+3^b-2^c\equiv 32\pmod 3\Leftrightarrow (-1)^a-(-1)^c=2\pmod 3\Leftrightarrow 2\mid a,2\not | c$. Let $\rm a=2d,c=2e+1$. If $\rm c\geq 3$ then $\rm 1+(-1)^b\equiv 0\pmod 4\Leftrightarrow 2\not |b$.Let $\rm b=2f+1$. $\rm 5^{2d}+3^{2f+1}-2^c\equiv 32\pmod 8\Leftrightarrow 25^d+3\cdot 9^f\equiv 0\pmod8 \Leftrightarrow 4\equiv 0\pmod8$,contradiction. So $\rm c=1$(because $\rm 2\not |c$).Therefore $\rm 5^a+3^b=34\Rightarrow a\leq 2$ $\rm a=1\Leftrightarrow 3^b=34-5=29$contradiction. $\rm a=2\Leftrightarrow 3^b=34-25=9\Leftrightarrow b=2$. So the only solution is $\boxed {\rm (a,b,c)=(2,2,1)}$.