Intro to the Quadratic Formula

Show that the equation $$(m^2+n)(n^2+m) = 2(m-n)^3$$ has no solutions in $(m, n)$ where $(m, n)$ are positive integers.

Try this!

Find the value of $$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\cdots}}}.$$

Let $$x = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\cdots}}}.$$ Then note that we actually have $$x = 1+\dfrac{1}{x}.$$ That means that $x^2 = x + 1$, or $x^2 - x - 1 = 0$.

To solve this, we can use the quadratic formula! The coefficients are $a = 1$, $b = -1$, $c = -1$, so the quadratic formula gives $$x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \dfrac{1\pm\sqrt{5}}{2}.$$

Something seems a bit fishy here. We have two values for $x$, but clearly it must be fixed to one value, right?

Actually, we're completely fine! If we look closely at the two possible values, we see that one of them ($\frac{1-\sqrt{5}}{2}$) is actually negative. Looking at the original question, we see that $x$ is definitely positive, so we can eliminate this value. That means that we have $$x = \boxed{\dfrac{1+\sqrt{5}}{2}}.$$

Note: This value is a special number known as the golden ratio, and is denoted by the greek letter $\phi$ (phi).