If $p$ is prime then $2p+1$ cannot be square How can I prove that $2p+1$ cannot be a square number if $p$ is prime? Is a contradiction proof enough where I assume true then show it as false eventually?
 A: Suppose $2p+1=k^2$. Clearly $p=2$ fails so we may assume $p$ is odd. 
We may rewrite it as $2p=(k-1)(k+1)$. Now since the LHS is even, so must the RHS. 
Note now that $k+1$ and $k-1$ must always have the same parity. Thus $k$ must be odd, so $k+1$ and $k-1$ are even. But this means that $(k+1)(k-1)$ must be divisible by $4$, and the LHS is clearly not divisible by 4, a contradiction.
A: (assuming $p>2$)
odd squares are all $1$ mod($4$), your expression is $3$ mod($4$).
A: If $2p+1$ is a square then $2p=n^2-1=(n-1)(n+1)$ for some odd number $n\ge 3$. Then $n-1$ and $n+1$ are even.
Now write 
$$p=\frac{n-1}2(n+1)$$
But then $p$ is an even number greater than $2$. Contradiction.
A: If $2p+1 = n^2$ then $2p = (n-1)(n+1)$, so then either $n-1$ or $n+1$ equal $2$, but neither  $0$ or $4$ are prime.
A: Assume its a square $\Rightarrow 2p + 1 = n^2$. If $n$ is even $\Rightarrow 2\mid n^2 \Rightarrow 2 \mid n^2 - 2p = 1$,  contradiction. If $n$ is odd $\Rightarrow n = 2k + 1 \Rightarrow 2p = n^2 - 1 = 4k(k+1) \Rightarrow p = k(k+1)$, a composite. Contradiction again, hence the conclusion .
A: It's clear for $1^2$. If $n>1$, we can express $n$ as $n=k+1$ for some $k$. If $(k+1)^2 = k^2 + 2k + 1 = 2z+1$ for $z$ prime we have $k^2 + 2k = 2z$. But then $k$ goes into $2z$, which contradicts $z$ being prime. Note the case with $k=2$ also yields a contradiction since $\frac{1}{2}(4 + 4)$ is not prime. 
