Team Selection Test

The answer is $(m,n,p)=(2,3,3)$, $(1,1,2)$ or $(2,2,5)$. Let $\frac{5^m+2^{n}p}{5^m-2^{n}p}=k^2$ for some positive integer $k$. Note that $5^m-2^{n}p\mid 5^m+2^{n}p$ implies that $5^m-2^{n}p\mid 2\cdot 5^m$. Then as $5^m-2^{n}p$ is odd, $5^m-2^{n}p\mid 5^m$ and hence $5^m-2^{n}p=5^r$ for some non-negative integer $r$.

Case 1: $r=0$ i.e. $5^m-2^{n}p=1$. If $n\ge 3$, then $5^m\equiv 1(\mathrm{mod}8)$ and hence $m=2s$ for some positive integer $s$. Then $5^{2s}\equiv 1(\mathrm{mod}3)$ and we have $2^{n}p\equiv 0(\mathrm{mod}3)$. Thus, $p=3$ and $(5^s-1)(5^s+1)=3\cdot 2^n$. Observe that $5^s+1\equiv 2(\mathrm{mod}4)$ and has an odd divisor greater than 3 when $s>1$.

Therefore $s=1$ and hence $m=2$, $n=3$ and $k=7$. If $n=2$, then $8p=(5^m+2^{2}p)-(5^m-2^{2}p)=k^2-1$. Therefore $k=2l+1$ for some positive integer $l$ and $2p=l(l+1)$. Then clearly $p=3$ and hence $5^m=13$ which yields a contradiction. If $n=1$, then $4p=(5^m+2^{1}p)-(5^m-2^{1}p)=k^2-1$. Therefore $k=2l+1$ for some positive integer $l$ and $p=l(l+1)$. Then clearly $l=1$, $p=2$ and hence $k=3$, $m=1$.

Case 2: $r\ge 1$. Then $5\mid 2^{n}p$ and hence $p=5$. Therefore, $5^{m-1}-2^n=5^{r-1}$ implies that $r=1$ since $m>r$ and $5^{r-1}\mid 2^n$. Thus, we have $5^{m-1}-2^n=1$. Clearly $n\ne 1$ and if $n=2$, then $m=2$ and $k=3$. If $n\ge 3$, then $5^{m-1}\equiv 1(\mathrm{mod}8)$ and hence $m-1=2s$ for some positive integer $s$. Then $(5^s-1)(5^s+1)=2^n$. Observe that $5^s+1\equiv 2(\mathrm{mod}4)$ and has an odd divisor greater than 1 when $s\ge 1$. Therefore $s=0$ and hence $2^n=0$ which yields a contradiction.