Show that
$$\int_{0}^{\infty} \frac{dx}{(1 + x^{\varphi})^{\varphi}} = 1$$
where $\varphi = \frac{\sqrt{5} + 1}{2}$ is the golden ratio.
Scroll down for solution:
Using the substitution $x^{-\varphi} = t$ and the identity $\varphi^2 = \varphi + 1$ we get that
$$dx = -\frac{1}{\varphi} t^{-\varphi} dt$$
Thus the integral becomes
$$\frac{1}{\varphi} \int_{0}^{\infty} \frac{dt}{(1 + t)^{\varphi}}$$
Which is easy to calculate.
Let $a_n$ be a sequence such that
$$a_1 = a \ge 2$$
and
$$a_{n+1} = a_{n}^2 - 2$$
Show that
$$\left(1 - \frac{1}{a_1}\right)\left(1 - \frac{1}{a_2}\right)\dots = \prod_{n \ge 1} \left(1 - \frac{1}{a_n}\right) = \frac{\sqrt{a^2 - 4}}{1 + a}$$
The putnam problem was with $a = \frac{5}{2}$
Scroll down for a solution.
Since $a \ge 2$, we can find a $u \ge 1$ such that $a = u + \frac{1}{u}$ (for $a = \frac{5}{2}, u =2$).
The case $a = 2, u = 1$ is easy, so assume $u \gt 1$.
The recurrence then gives us
$$a_{n+1} = u^{2^n} + \frac{1}{u^{2^n}}$$
Let $$f(x) = \left(1 - \frac{1}{x + \frac{1}{x}}\right) = \frac{x^2 - x + 1}{x^2 + 1}$$
The value we are seeking is
$$f(u)f(u^2)f(u^4)f(u^8)\dots f(u^{2^n})\dots$$
Now notice that
$$(x^2 - x + 1)(x^2 + x + 1) = (x^2 + 1)^2 - x^2 = x^4 + x^2 + 1$$
Thus if we multiply
$$(x^2 - x +1)(x^4 - x^2 + 1)(x^8 - x^4 + 1)\dots(x^{2^{n}} - x^{2^{n-1}} + 1) $$
with $x^2 + x + 1$ we get
$$(x^2 + x + 1)(x^2 - x +1)(x^4 - x^2 + 1)(x^8 - x^4 + 1)\dots(x^{2^{n}} - x^{2^{n-1}} + 1) = $$
$$(x^4 + x^2 + 1)(x^4 - x^2 + 1)\dots(x^{2^{n}} - x^{2^{n-1}} + 1) = x^{2^{n+1}} + x^{2^n} + 1$$
and similarly using
$$(x^2 - 1)(x^2 + 1) = x^4 - 1$$
we get
$$(x^2 - 1)(x^2 + 1)(x^4 + 1) \dots (x^{2^n} + 1) = x^{2^{n+1}} - 1$$
Since
$$\frac{x^{2^{n+1}} + x^{2^n} + 1}{x^{2^{n+1}} - 1} \to 1$$
(for $x >1$)
We see that
$$L = f(u)f(u^2)\dots f(u^{2^n})\dots = \frac{u^2 - 1}{u^2 + u + 1}$$
This we can rewrite as
$$ L = \frac{u - \frac{1}{u}}{u + \frac{1}{u} + 1} $$
Squaring gives
$$L^2 = \frac{u^2 + \frac{1}{u^2} - 2}{(a+1)^2} = \frac{a^2 -4}{(a+1)^2}$$
Thus
$$ L = \frac{\sqrt{a^2 - 4}}{a+1}$$
