Problem 2

The equations are $2b-c=2^m$, $2c-b=2^n$, and $bc-2=2^p$. Since $n\ge1$, the second equation makes $b$ even. If $p=1$, then $bc=4$; with $2\le b\le c$ this gives $(b,c)=(2,2)$. Now suppose $p>1$. Since $bc=2^p+2\equiv2\pmod4$ and $b$ is even, $b\equiv2\pmod4$ and $c$ is odd. Thus $2b-c$ is odd, so $m=0$ and $2b-c=1$, i.e. $c=2b-1$. Substitution into $2c-b=2^n$ gives $3b=2^n+2$ and hence $3c=2^{n+1}+1$. Also $bc-2=2^p$ becomes $(2^{n-1}+1)(2^{n+1}+1)=9(2^{p-1}+1)$. Here $p\ge n$; reducing this identity modulo $2^{n-1}$ gives $1\equiv9\pmod{2^{n-1}}$, so $n\le4$. The case $n=1$ is impossible in $3b=2^n+2$, and $n=3$ is impossible because $b=(2^n+2)/3$ is not an integer. Thus $n=2$ or $4$, yielding $(b,c)=(2,3)$ or $(6,11)$.