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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?

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    $\begingroup$ Is $q$ even or odd? $\endgroup$ – Jyrki Lahtonen Jun 21 '17 at 19:39
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Hint Since $p <q$ are prime numbers, then $q \geq 3$ and hence $q$ is odd.

Since $2p$ is even and $q$ is odd, $2p+q^2$ is ....

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  • $\begingroup$ This is perfect. Thank you for helping my brain process it this way. $\endgroup$ – 1011011010010100011 Jun 21 '17 at 19:41
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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.

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Thank you very much @N.S.

Since $\ p$ and $\ q$ are prime numbers and $\ p<q$, then $\ q>= 3$ (because the lowest and only even prime is 2), and hence q is always odd. Since $\ 2p$ is an even integer and q is odd, then by (1a) $\ 2p+q^2$ is odd.

Clause (1a): Prove that if $\ a$ is even and $\ b$ is odd, then $\ a+b$ is odd.

If $\ a$ is even then $\ a = 2k$ for some integer $\ k$. Is $\ b$ is odd then $\ b = 2j+1$ for some integer $\ j$. Then, by substitution, $\ a+b=2k+2j+1 = 2(k+j)+1 = 2b+1$ where $\ b$ is an integer and $\ b=k+j$. Hence, the addition of an even and an odd is always odd.

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  • $\begingroup$ You can almost certainly just assert that even+odd=odd. It's something that almost everybody will agree is so obvious that it doesn't need proof. $\endgroup$ – David Richerby Jun 21 '17 at 22:09

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