If $m$ is even, and $n$ is odd, does $2(m+n)+2$ have to be divisible by $4?

Question

Can anybody give me an idea of how to solve this? I can't seem to find a counterexample because every integer I choose for m and n is divisible by 4.

Answer 1

The proof is flawed, but the theorem is in fact correct, so you won’t be able to find a counterexample to it.

The error in the proof is using $k$ in the expansions of both $m$ and $n$, thereby making $n=m+1$. The correct argument is that there are integers $k$ and $\ell$ such that $m=2k$ and $n=2\ell+1$, so that

$$\begin{align*} 2(m+n)+2&=2\big(2k+(2\ell+1)\big)+2\\ &=4k+2(2\ell+1)+2\\ &=4k+4\ell+2+2\\ &=4(k+\ell+1)\;, \end{align*}$$

which is indeed a multiple of $4$.

Answer 2

The error is that assuming m is even and thus equal to 2k for some k, and that in assuming that n is odd and thus equal to 2k + 1 for some k, that these are the same value when they need not be.

This leads to some false assumptions. For example: n and m are consecutive, which of course they don't need to be. Or that m + n = 2k + 2k + 1 = 4k + 1, is 1 more than a multiple of 4. Neither of these need to be true.

BUT then this gets weird. The proof that 2(m+n) + 2 is divisible by 4 is a true result. We can't do 2(m + n) + 2 = 2(4k + 1) + 2. But we can (and must do 2(m + n) + 2 = 2(2k + 2j + 1) + 2 = 4(k+j) + 2 + 2 = 4(k+j) + 4= 4(k+j + 1).

So 2(m+n) +2 is always divisible by 4.

However, this would be a better example of a false application leads to untrue theorem:

Prove 2(m + n) + 2; m even and n odd is never divisible by 8.

Proof:

m = 2k

n = 2k + 1

so 2(m + n) + 2 = 2(4k + 1) + 2 = 8k + 4 and is not divisible by 8.

This is of course not true.

m = 2k

n = 2j + 1

so 2(m + n) + 2 = 2(2(j + k) +1) + 2 = 4(j+k) + 4 which will be divisible by 8 if (j + k) is odd.

Example: if m = 4 = 2*2 and n = 7 = 2*3 + 1 so 2(4 + 7) + 2 = 24 = 3*8.