How do I prove or disprove this prime number conjecture? How would I go about proving something like this? 
Prove or disprove: If $p$ and $q$ are prime numbers for which $\ p < q$, then $\ 2p+q^2$ is odd. 
I'm assuming its definitely true because and even $+$ odd is always odd, and odd $\times$ odd is always odd too. So I know it's true, but how do I prove it? 

Since you know that $2p$ is even, this is true iff $q^2$ is odd for all $p$ and $q$ which satisfy that condition. Since you can't choose the only even prime to be $q$, as there are no primes less than it to take the role of $p$, you know $q^2$ to be odd, and you are done.
