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How can we prove that if [math]r + \frac{1}{r}[/math] is an integer, then [math]r^n + \frac{1}{r^n}[/math] is also an integer?
Let’s play around. [math]r^2 + \dfrac{1}{r^2}[/math] is like [math]\left( r + \dfrac{1}{r} \right)^2[/math], except that expanding the square actually gives [math]r^2 + 2 + \dfrac{1}{r^2}[/math]. Fortunately, 2 is an integer, and therefore, [math]r^2 + \dfrac{1}{r^2} = \left( r + \dfrac{1}{r} \right)^2 - 2[/math] is the difference of two integers.Now on to [math]n = 3[/math]. That would of course be like [math]\left(r + \dfrac{1}{r}\right) \left(r^2 + \dfrac{1}{r^2}\right)[/math], except again for some cross terms. The actual product is:[math]\displaystyle \left(r + \dfrac{1}{r}\right) \left(r^2 + \dfrac{1}{r^2}\right) = r^3 + r + \frac{1}{r} + \frac{1}{r^3},[/math]and therefore[math]\displaystyle r^3 + \frac{1}{r^3} = \left(r + \dfrac{1}{r}\right) \left(r^2 + \dfrac{1}{r^2}\right) - \left(r + \dfrac{1}{r}\right),[/math]again the difference of two integers. The general case should, I hope, now be obvious and the proof by induction a straightforward extension of the above.
Is [math]10^{100}+1[/math] (a googol plus 1) prime?
I will try to give a simpler perspective that what many other people have given here.Remember the Good old finite GP[math]1+r+r^2+....+r^{n-1}=\frac{r^n-1}{r-1}[/math]?This simply means that if r is replaced by -r, we get[math]1-r+r^2-r^3+r^4...r^{n-1}=\frac{{(-r)}^n-1}{-r-1}[/math]Thus, [math]1-r+r^2-r^3+r^4...+r^{n-1}=\frac{1-{(-r)}^n}{r+1}[/math]If n is an odd number,Thus, [math]1-r+r^2-r^3+r^4...+r^{n-1}=\frac{1+r^n}{r+1}[/math]Thus, If n is an odd number,[math]1+r^n=(1-r+r^2-r^3+r^4...r^{n-1})(r+1)[/math]Thus [math]1+r[/math] is a factor of [math]1+r^n[/math] each time n is an odd number.Now lets see our question.[math]10^{100}+1 = {(10^{20})}^5+1[/math]Thus [math]10^{20}+1[/math] is a factor of [math]10^{100}+1[/math]
How do you evaluate [math]\displaystyle\prod_{r=0}^{\infty}\dfrac{r³-1}{r³+1}[/math] ?
As observed, we have a factor of [math]0[/math] in this product (the [math]r = 1[/math] factor); hence we obtain[math]\displaystyle \prod_{r=0}^{\infty} \frac{r^3 - 1}{r^3 + 1} = 0. \tag*{}[/math]This product is more interesting if we start with [math]r = 2[/math]; let[math]P = \displaystyle \prod_{r=2}^{\infty} \frac{r^3 - 1}{r^3 + 1}. \tag*{}[/math]By the sum and difference of cubes formulas, we can write[math]\begin{align*} P &= \displaystyle \lim_{n \to \infty} \prod_{r=2}^n \frac{r - 1}{r + 1} \cdot \frac{r^2 + r + 1}{r^2 - r + 1}\\ &= \displaystyle \lim_{n \to \infty} \prod_{r=2}^n \frac{r - 1}{r + 1} \cdot \prod_{r=2}^n \frac{r(r+1) + 1}{(r-1)r + 1}. \end{align*} \tag*{}[/math]This yields a pair of telescoping products:[math]\begin{align*} P &= \displaystyle \lim_{n \to \infty} \Big(\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot ... \cdot \frac{n - 1}{n + 1} \Big) \cdot \Big(\frac{2(3)+1}{1(2)+1} \cdot \frac{3(4)+1}{2(3)+1} \cdot ... \cdot\frac{n(n+1) + 1}{(n-1)n + 1}\Big) \\ &= \displaystyle \lim_{n \to \infty} \frac{1 \cdot 2}{n(n + 1)} \cdot \frac{n(n+1) + 1}{1(2)+1}\\ &= \frac{2}{3}. \end{align*} \tag*{}[/math]Therefore, we conclude that[math]\displaystyle \prod_{r=2}^{\infty} \frac{r^3 - 1}{r^3 + 1} = \frac{2}{3}. \tag*{}[/math]
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