If $x$ is an even integer, then $x^2 - 6x + 5$ is odd.
My solution (direct proving): $$ x = 2k$$ $$ x^2 - 6x + 5 = 4k^2 -12k + 5 $$ $$ 4k^2 -12k + 4 + 1 = 2(2k^2-6k+2)+1$$ which is by definition is odd. Is my solution correct?
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3yes. Also, $x^2-6x+5=(x-1)(x-5)$ both factors are odd – lab bhattacharjee Jun 06 '13 at 15:19
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how can you tell? can you use it in a direct proof? – jantristanmilan Jun 06 '13 at 15:21
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5even - odd =odd and odd*odd=odd – lab bhattacharjee Jun 06 '13 at 15:23
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But the fact that @vincentbelkin is showing such a statement probably means he shouldn't cite that unless he proves it as well. His way is more direct unless he can take odd*odd = odd as given. – Suugaku Jun 06 '13 at 15:24
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It's excellent. Any integer of the form $2m+1$ is odd. You just need to say that your original $k$ is an integer and so, too, is $2k^2-6k+2$.
Ted Shifrin
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Very nicely done: When you use the premise that $x$ is an even integer, so that $x = 2k$, you need only add:
..."for some $k \in \mathbb Z$. And so $4k^2 -12k + 4 + 1 = 2(2k^2-6k+2)+1$ is an odd integer."
Note also (alternatively), that, if $x$ is even, then there exists a $k\in \mathbb Z$ such that $x = 2k$. $$x = 2k \implies x^2-6x+5=(x-1)(x-5) \implies (2k - 1)(2k-5)\;\;\text{ is necessarily odd}.$$
I do think that your route to proving this is the most direct - just don't be afraid to use a few words here and there, where appropriate!
amWhy
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Yes, the proof is correct. The same idea shows that if $\,f(x)\,$ is a polynomial in $\,x\,$ with integer coefficients then, when $\,x\,$ is even, $\,f(x)\,$ has the same parity as its constant coefficient $\, = f(0),\,$ and when $\,x\,$ is odd, $\,f(x)\,$ has the same parity as the sum of the coefficients. This will be clearer arithmetically when one learns about modular arithmetic. Then it can be expressed as
$$ f(x) = f_0 +\cdots + f_n x^n\ \Rightarrow\ {\rm mod}\ 2\!:\, \bigg\lbrace\,\begin{eqnarray} x\equiv 0\ &\Rightarrow&\, f(x)\equiv f(0) \equiv f_0\\ x\equiv 1 \ &\Rightarrow&\, f(x)\equiv f(1)\equiv f_0+\cdots + f_n\end{eqnarray}$$
If particular, if the above values are $\,\rm\color{#c00}{both\equiv 1},\,$ then $\,f\,$ takes only odd values, so it does not take the even value $\,0,\,$ therefore $\,f\,$ has no integer roots. Hence we have deduced the following simple
Theorem $\ $(Parity Root Test) $\ $ A polynomial $\,f(x)$ with integer coefficients has no integer roots when its constant coefficient and coefficient sum are $\rm\color{#c00}{both\ odd}$.
