What would be a proof strategy for this? I would like to show a proof of the contrapositive: if the expression is not odd, then $x$ is not an integer. If I go that route, how do I express the expression as odd in algebraic terms?
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8Since $x(x+5)$ is even(why?), $x^2+5x-1=x(x+5)-1$ is odd. – mathlove Apr 29 '15 at 22:16
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1Thank you! it's clear now. – kvax12v Apr 29 '15 at 22:19
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$n$ is odd if there exists $k\in\Bbb N $ such that $n=2k+1$ and $n$ is even if there exists $l\in \Bbb N$ such that $n=2l$. – Surb Apr 29 '15 at 22:19
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Got it! Thanks! – kvax12v Apr 29 '15 at 22:21
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2Yet another approach: $x^2+5x-1=x^2+x+4x-1=\color{red}{x(x+1)}+\color{blue}{(4x-1)}$. Now $\color{red}{x(x+1)}$ is always Even as product of two consequent numbers implies one of them has be to be even. $\color{blue}{(4x-1)}$ is always Odd as it is of the form $2k-1$. Hence we have $$x^2+5x-1=\text{Even}+\text{Odd}=\text{Odd}$$ – Mufasa Apr 29 '15 at 22:27
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1I really like that approach, thanks mufasa! – kvax12v Apr 29 '15 at 22:30
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See the Parity Root Test – Bill Dubuque Apr 29 '15 at 22:39
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2Use modulo $2$: $x^2\equiv x$, $5x\equiv x$, $-1\equiv 1$, so $x^2+5x-1\equiv x+x+1\equiv 2x+1\equiv 1\pmod{2}$. – egreg Apr 29 '15 at 23:05
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I will go the direct way
You have two cases.
case (1)
$x$ is even then $x = 2k$ for some integer $k$ and so $$x^2 + 5x - 1 = (2k)^2 + 5(2k) - 1 = 4k^2 + 10k - 1 = 2(2k^2 + 5k) - 1$$which is an odd number because $2k^2 + 5k$ is an integer , maybe call it $m$ and $2m-1$ is an odd integer.
case(2)
$x$ is odd then $x= 2k+1$ for some integer $k$ and so $$x^2 + 5x - 1 = (2k+1)^2 +5(2k+1) - 1 = 4k^2 + 4k + 1 + 10k + 5 -1 = 4k^2 + 14k + = 2(2k^2 + 7k) + 5$$
Now it should be obvious to you that this is also an odd integer because $2k^2 +7k$ is an integer, call it $n$ and so $2n + 5 = 2(n+3) -1$ is odd and so you are done !
Remember that an any integer has the form of $2k+1$ or $2k-1$ and any even integer has the form of $2k$ where $k$ is an integer of course.
alkabary
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Hint: Write polynomials as binomial coefficients (combinatorial polynomials). $$ x^2+5x-1=2\left[\binom{x}{2}+3\binom{x}{1}\right]-1 $$
robjohn
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