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I've been asked to show why an equation has no complex roots but i'm at a complete loss.

The equation is

$F_{n+2}=F_n$

Where $F_n=(x-1)(x-2)...(x-n)$ and n is a positive integer.

I'd really appreciate if someone could explain how I could go about showing this because I'd really like to understand.

Thanks in advance.

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5  
What is intended is clear, but every real number is a complex number, so every root is complex! It really should read show that every root is real. –  André Nicolas Aug 24 at 5:59
    
The result holds for $F_n=(x-a_1)(x-a_2)\cdots(x-a_n)$, for every real valued sequence $(a_n)$. Solving this more general case may help to shed light on the specific case at hand. –  Did Aug 24 at 8:10

3 Answers 3

$F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$.

So, the equation $F_{n+2}=F_n$ definitely has $1,2, \cdots n$ as the roots. Now "cancel" out the common terms. You are left with the equation:

$(x-(n+1))(x-(n+2))=1$. Solve this quadratic directly to show that there are no complex roots.

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Thanks allot!. I'll look into this! –  Like to understand Aug 24 at 4:32
    
There are more roots of $F_{n+2} = F_n$other that those of your last equation. Any root of $F_n = 0$ would be a root of $F_{n+2} = F_n$. –  leo Aug 24 at 4:34
2  
@leo: and roots of $F_n=0$ are...? –  voldemort Aug 24 at 4:35
1  
@leo: and those roots were mentioned in my answer- please read the whole text :) –  voldemort Aug 24 at 4:44
2  
@Liketounderstand BTW we know that $(x-(n+1))(x-(n+2))-1=0$ has two real roots even without explicitly solving the equation. It suffices to notice that for $x\to \pm\infty$ we get positive values, and negative values are also possible, for example for $x=n+1$. –  Martin Sleziak Aug 24 at 8:07

If $$F_n=(x-1)(x-2)...(x-n)$$ then $$F_{n+2}=(x-1)(x-2)...(x-n)(x-(n+1)) (x-(n+2))$$ So, if $F_n=F_{n+2}$, you have $$(x-1)(x-2)...(x-n)=(x-1)(x-2)...(x-n)(x-(n+1)) (x-(n+2))$$ So, you can factor and arrive to $$\Big((x-(n+1)) (x-(n+2)) -1\Big)F_n=0$$ The roots of the equation are those of $F_n=0$ (that is to say $1,2,3,\cdots,n-1,n$) and the roots of the factor $$(x-(n+1)) (x-(n+2)) -1$$ Expanding and grouping terms, we arrive to the following quadratic $$x^2-(2 n+3) x+(n^2+3 n+1)=0$$

I am sure that you can take from here.

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$F_{n}=(x-1)(x-2) \cdots (x-n)$

$F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$

$F_{n} = F_{n+2}$

$(x-1)(x-2) \cdots (x-n)=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$

$(x-1)(x-2) \cdots (x-n)[(x-(n+1))(x-(n+2)) - 1] = 0$

$(x-1)(x-2) \cdots (x-n)[(x^2-[(n+2)+(n+1)]x+(n+1)(n+2) - 1] = 0$

$(x-1)(x-2) \cdots (x-n)[x^2-(2n+3)x+n^2+3n+ 1] = 0$

Therefore $x = 1,2,3,...,(n-1),n$ and $x^2-(2n+3)x+n^2+3n+ 1 = 0$

$$x=\frac{(2n+3)\pm \sqrt{(2n+3)^2-4(1)(n^2+3n+1)}}{2(1)}$$

$$x=\frac{(2n+3)\pm \sqrt{5}}{2}$$

$$x=\frac{2n+3+ \sqrt{5}}{2} x=\frac{2n+3- \sqrt{5}}{2}$$

Therefore the roots are not complex

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Two other answers already explained that, didn't they? (Only, they abstained (wisely) to fully complete the computations.) –  Did Aug 24 at 8:07

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