I need to prove that if both $ m$ and $n$ are odd integers, then $mn+2$ is odd. I found several similar answers for a problem that asks if $m$ and $n$ are odd the prove that $mn$ is odd as well. This question adds an even number and wants to prove that it is odd.
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$\begingroup$ Welcome to Mathematics Stack Exchange. Odd + Even = Odd $\endgroup$ – J. W. Tanner Oct 27 '20 at 2:05
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$\begingroup$ Break it into steps: first show that $mn$ is odd, then that $mn+2$ is also. $\endgroup$ – abiessu Oct 27 '20 at 2:07
If $m=2k+1$ and $n=2l+1$ for some integers $k$ and $l$,
then $mn+2=4kl+2k+2l+3=2(2kl+k+l+1)+1$.
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If $m$ and $n$ are odes then $m+1, n+1$ and $m+n$ are even.
In particular, $(m+1)(n+1)-(m+n)$ is even. But $mn+2=(m+1)(n+1)-(m+n)+1$ is odd.
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If you know that $m$ is odd then you can say that it is in some way equal to $(2p+1)$. Likewise, $n$ is also equal to $(2p+1)$. Then multiply these together to get $mn=4p^2+4p+1$. This must be odd because we can set $4p^2+4p+1=2(2p^2+2p)+1$ and if $(2p^2+2p)$ is set equal to $b$ then $mn=2b+1$ which is an odd number. Adding $2$ does not change our answer because if $mn=4p^2+4p+1$ and we add $2$, $mn+2=4p^2+4p+1+2$, then we can say that if $g=(2p^2+2p+1)$ then $mn+2=2(2p^2+2p+1)+1=2g+1$ which is odd.