I have another question, On the prime divisor of $n=(p^{2}1)/2$, where $p$ is prime, please tell me your idea. Let $p\neq 3$ be Merssen prime. Is it true $n$ has a prime divisor $r$ such that $r^{2}$ does not divide $n$?

Well, maybe I'm missing the point, but the first Mersenne prime is $3=2^21$, and $\frac{3^21}{2} = 4 = 2^2$, so... no, I guess. 


Considering there are only 47 known Mersenne primes, and finding the factors of $2^p1$ is a difficult task, I'm not sure this question is fully tractable. But we can certainly show that all of the known Mersenne primes satisfy your question. First, take $m=2^p1$ to be a Mersenne number, and rewrite $n= \frac{m^2  1}{2} = 2^{2p1}  2^p$. This is powerful when $p=2$, which gives Philip van Reeuwijk's counterexample. For larger $p$, we know $4$ will always divide $n$, so we can ignore powers of $2$ and just consider the odd part $n'= 2^{p1}  1$. We know $p$ is prime, and odd since it is not $2$, so we are looking at $n' = 4^k  1$, with $k=\frac{p1}{2}$. Clearly $3$ divides $n'$, and $9$ divides iff $3$ divides $k$. So we require that $p \equiv 1 \mod6$. This is not enough; plenty of known Mersenne primes have this property. Since $3$ divides $k$, we can write $n'=64^{k'}1$. Then $7$ divides $n'$, and $7^2$ divides $n'$ iff $7$ divides $k'$. So now we require that $p \equiv 1 \mod42$. Sadly, again this is not enough. One step further, we see that when $7$ divides $k'$, $43$ divides $64^{k'}1$, and $43^2$ divides iff $43$ divides $k'$. Now we're happy (for the time being), because no known Mersenne prime has $p\equiv 1 \mod 1806 = 43*7*6$. But it seems there's no reason they can't have this property, so you may have to continue your search once such a Mersenne prime is found. Expect one by the 504th instance: $504 = \phi(1806)$. 

