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If $p$ and $q$ are prime numbers for which $p < q$, then $2p+q^{2}$ is odd. Suppose $p$ and $q$ are prime and $p<q$. Thus, by definition of prime, 2 is the only even prime number. Case 1: Suppose $p=2$. Since, $2<q$ and q is prime. It follows that q is odd.Thus, q=2k+1 for some $k\in\mathbb{Z}$. Thus, \begin{align*} 2p+q^{2}&=2(2)+(2k+1)^{2}\\ &=4+4k^{2}+4k+1\\ &=2(2k^{2}+2k+2)+1\\ \end{align*} Thus, by closure of the set of integers under integer addition and multiplication,$2k^{2}+2k+2\in\mathbb{Z}$. Thus,$2p+q^{2}=2m+1$ for some $m\in\mathbb{Z}$, namely, $m=2k^{2}+2k+2$. Thus, $2p+q^{2}$ is odd.

Case 2: Suppose $p\neq 2$. Thus $p$ is odd and $q$ is odd.Thus $p=2k+1$ and $q=2l+1$ for some $k,l\in\mathbb{Z}$. Thus, \begin{align*} 2p+q^{2}&=2(2k+1)+(2l+1)^{2}\\ &=4k+2+4l^{2}+4l+1\\ &=2(2k+2l^{2}+2l+1)+1. \end{align*} Thus, by closure of the set of integers under integer addition and multiplication, $2k+2l^{2}+2l+1\in\mathbb{Z}$. Thus, $2p+q^{2}=2m+1$ for some $m\in\mathbb{Z}$, namely, $m=2k+2l^{2}+2l+1$. Thus, $2p+q^{2}$ is odd. Therefore, if $p$ and $q$ are prime numbers for and $p < q$, $2p+q^{2}$ is odd. .

Any problems with my solution?

HighSchool15
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    I don't see any problems, but you didn't really need to split it into two cases. Regardless of whether $p = 2$ or not, we must have that $q$ (and hence $q^2$) is odd, and regardless of whether $p=2$ or not, we must have that $2p$ is even. – Joey Zou Oct 21 '16 at 04:36
  • No problems...Absolutely fine!!:-) – Soham Oct 21 '16 at 04:58
  • This proof is ok, but is unnecessary long. You should note that if $p<q$, then $q$ must be at least $3$, so $q^2$ is odd and then, $2p+q^2$ is an even number plus an odd number, hence an odd number. – Xam Oct 21 '16 at 20:08

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