, 2015

Let $\frac{5^a + 2^b p}{5^a - 2^b p} = n^2$. Since $5^a - 2^b p$ is odd, we also have $5^a + 2^b p$ is even. Thus $5^a - 2^b p$ and $5^a + 2^b p$ are both positive integers. However, the number $5^a - 2^b p$ is odd, hence $5^a - 2^b p$ is odd. Thus we can write $5^a - 2^b p = 2 \cdot 5^a$. However, the number $5^a - 2^b p$ is odd, hence $5^a - 2^b p$ is odd. We consider two cases:
* If $m \ge 1$, we have $2^b p = 5^a - 5^m$, hence $2^b p$ is divisible by $5$. It follows $p=5$ and we get equation $2^b = 5^{a-1} - 5^{m-1}$. Since $2^b$ is not divisible by $5$ the same holds for $5^{m-1}$. Thus the only possibility is $m=1$ and we get equation
$$5^{a-1} - 2^b = 1. \qquad (1)$$
Obviously, $b=1$ is not a solution, and if $b=2$ we get a solution $a=2$. For $b \ge 3$ view the equation mod $8$. We have $2^b \equiv 0 \pmod 8$. Hence $5^{a-1} \equiv 1 \pmod 8$, which means that $a-1=2k$ for some $k \in \mathbb{N}$. We can rearrange the equation (1) to $(5^k-1)(5^k+1) = 2^b$. Therefore the numbers $5^k-1$ and $5^k+1$ are both powers of $2$. The only powers of $2$ that differ by $2$ are $2^1$ and $2^2$. This would mean that $5^k=3$ which is not possible.

The only solution of the equation (1) is thus $a=2$ and $b=2$. Inserting this in the original equation we get $n=3$. Thus we get a solution $(2,2,5)$.

$$5^a - 2^b = 1.$$
If $b=1$ we get $5^a - 1 = 2p$. Since $4|5^a - 1$ the only possibility is $p=2$ and $a=1$. Inserting this in the original equation we get $n=3$. Thus $(1,1,2)$ is a solution.
If $b = 2$ we get $5^a - 1 = 4p$. From the original equation then follows $n^2 = 8p+1$ which gives $(n+1)(n-1) = 8p$. The number $n$ must be odd and bigger than $1$, hence we write $n = 2s + 1$. We get $(s+1)s = 2p$. The only solution of this equation is $s = 2$ and $p = 3$. However, this would mean that $5^a = 13$, which has no solutions in positive integers.
We are left with the case $b \ge 3$. In this case we have $2^b p \equiv 0 \pmod 8$ and hence $5^a \equiv 1 \pmod 8$. Thus $a = 2k$. However, we then get $5^{2k} \equiv 1 \pmod 3$, hence $2^b p \equiv 0 \pmod 3$ must hold, which implies $p = 3$. It follows that $(5^k - 1)(5^k + 1) = 3 \cdot 2^b$. However $5^k + 1 \equiv 2 \pmod 4$, hence $5^k + 1$ must have some odd divisor. The only possibility is thus $5^k + 1 = 2 \cdot 3$. This gives $a = 2$ and $b = 3$. Inserting these values in the original equation we get $n = 7$. Thus $(2, 3, 3)$ is a solution.

Therefore the solutions are $(1, 1, 2)$, $(2, 3, 3)$, and $(2, 2, 5)$.