Show by induction $n^2 - 3n + 4$ is even I have this question I am working on. I am not sure If I am doing the correct thing.
Show by induction $n^2-3n+4$ is even.

*

*Basis step
$n = 1$ $\implies$$1^2 - 3(1) + 4 = 2$


*$n^2 - 3n + 4$ is even means


*$n^2 - 3n + 4 = 2m$ for some $m$


*Induction step $n=k$


*$k^2 - 3k + 4=2m$


*$(k+1)^2 - 3(k+1) + 4 = 2m$


*From step 5, $k^2 - 3k = 2m - 4$


*$\frac{k^2 - 3k} {2} = m - 2$


*Which comes to $0=0$ when $n=1$


*Now I want to prove the same for $k+1$


*From step 6, $(k+1)^2 - 3(k+1) = 2m - 4$


*$\frac{(k+1)^2 - 3(k+1)} {2} = m - 2$


*Which is $0=0$ when $k=2$
What I have done above, is the best I could come up with but I know it's wrong somehow. Can you correct me if there are any mistakes and if there is somewhere I am going wrong.
 A: *

*The $\implies$ sign makes no sense here.

*Right.

*Right.

*Right.

*Right.

*Wrong: $(k+1)^2-3(k+1)+4=2n$ or $2j$ or $2l$. Any letter but $m$ (or $k$), since $m$ is the natural number such that $k^2-3k+4=2m$.

*Right.

*Right.

*This makes no sense, since there is no $n$ here.

And neither the rest makes sense.
Note that$$\bigl((k+1)^2-3(k+1)+4\bigr)-(k^2-3k+4)=2k-2,$$which is an even number. So, since $(k+1)^2-3(k+1)+4$ is equal to $k^2-3k+4$ plus an even number, if you assume that $k^2-3k+4$ is even, then so will be $(k+1)^2-3(k+1)+4$.
A: To prove something by Induction we must:

*

*Show that it is true for the smallest value on the set we are proving it

*Suppose it is true for $k$

*Show it is true for $k+1$

*Then say it is therefore true for all $n$
I assume $n \in \mathbb{Z}^+$ meaning that our smallest value is $0$. So,
$$P_0 = 0^2 -3\cdot0 + 4 = 2\cdot 2 $$
Meaning that this is true for $P_0$
We now assume that $P_k$ is true meaning that:
$$ P_k = k^2 -3k + 4 = 2a\, , \quad a \in \mathbb{Z}^+$$
We now want to show that $P_{k+1}$ is true:
$$ P_{k+1} = (k+1)^2 -3(k+1) + 4 = 2b \, , \quad b \in \mathbb{Z}^+$$
Expanding:
$$ P_{k+1} = k^2 + 2k + 1 -3k -3 + 4 = k^2 -k + 2$$
We can now make a substitution. Remember that
$$ P_k = k^2 -3k + 4 = 2a$$
so
$$ k^2 = 2a +3k -4$$
Plugging this in $P_{k+1}$
$$2a + 3k -4 -k + 2 = 2a + 2k -2 = 2(a+ k -1) $$
$P_{k+1}$ was set to be equal to $2b$. We have now shown that $P_{k+1}$ is true meaning that by induction $P_n$ is true.
A: We can make the induction step much shorter:
Set $P(k)=k^2-3k+4$. Then $\:\Delta P(k)=P(k+1)-P(k)=\dots= 2(n-1)$, so $P(k+1)$ and $P(k)$ have the same parity, and the induction step follows.
