How Are They Similar?

$A,B$ are $19,15$ respectively(taken when $m=1, n=2$ ), giving the answer $19\cdot 15=\boxed {285}$ . To see why, we prove a more general result: Given positive integer $p$ , the smallest positive integer that can be expressed as $(2p)^{2m}-(2p-1)^n$ is $4p-1$ , where $m,n$ are still positive integers.

Note that $p=5,4$ respectively for our original problem.

---

We consider two cases for $n$ :

If $n$ is even, then $\begin{aligned} (2p)^{2m}-(2p-1)^n &=(2p)^{2m}-(2p-1)^{2(\frac {n}{2})}) \\ &=((2p)^m-(2p-1)^{\frac {n}{2}})((2p)^m+(2p-1)^{\frac {n}{2}}) \\ &\ge 2p+2p-1 \\ &=4p-1 \end{aligned}$ The inequality is valid since $(2p)^m-(2p-1)^{\frac {n}{2}}>0$ because we want a positive number and $m, \frac {n}{2}$ are positive integers. Thus, equality holds when $m=1, n=2$ .

If $n$ is odd, then consider the modulo $4p$ of $(2p)^{2m}-(2p-1)^n$ . Clearly $(2p)^{2m}\equiv 0 \pmod {4p}$ , while $\begin{aligned} (2p-1)^n &= (2p)^n-n(2p)^{n-1}+...+n(2p)-1 \\ &\equiv n(2p)-1 \\ &\equiv 2p-1 \pmod {4p} \end{aligned}$ by binomial expansion and $(2p)^k\equiv 0 \pmod {4p}$ for $k>1$ .

This means that $(2p)^{2m}-(2p-1)^n \ge 2p-1$ . However, the equality clearly cannot hold since $2p-1|(2p)^{2m}$ but $(2p-1,2p)=1$ . Hence $(2p)^{2m}-(2p-1)^n \ge 4p+2p-1=6p-1$ , which is greater than the lower bound we found for $n$ is even. Thus we have proven our proposition.