Find a Defining Equation for the Golden Ratio: $(1 + \sqrt{5})/2$ Find a defining equation for the golden ratio: $(1 + \sqrt{5})/2$. Furthermore, find its norm in $\mathbb{Q}[\sqrt{5}]$. It seems like it would be simple, but I am still new to quadratic field. Could I get some aid on how to start and solve this? Thanks!
 A: $$x= (1 + √5)/2$$
$$2x= 1 + √5$$
$$2x-1=  √5$$
$$4x^2-4x +1=  5$$
$$4x^2-4x -4=  0$$
$$x^2-x -1= 0$$
gives one possibility, though the squaring has introduced another root, namely $x= (1 - √5)/2$.  If you multiply these two roots together you will get $-1$, the norm.  See https://en.wikipedia.org/wiki/Field_norm for more on finding norms
A: $\!\!\!\begin{align}{\rm Recall}\ \ w = a + b\sqrt{n}\rm \ \ has\ \ {\bf norm}\  &=\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2\! - n\: b^2\\[4pt]
{\rm and,\ furthermore,\ }w\rm \ \ has\ \ {\bf trace}\ &=\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2a\end{align}$
Now apply Vieta $\ (x\!-\!w)(x\!-\!w') = x^2 - (w\!+\!w')x + ww' =\, x^2 - 2a\, x + a^2\!-nb^2$
A: The golden ratio is one of the solutions to $x^2-x-1=0$. (I presume this is what you mean by "defining equation.")
Its norm would be $\left(\frac{1}{2}\right)^2 - 5 \cdot \left(\frac{1}{2}\right)^2 = -1$.
A: Suppose you split a length in two. You split it according to the golden ratio if the ratio of the larger length $L$ to the smaller length $l$ is equal to the ratio of the total length $L+l$ to the larger length $L$. 
In formulae, if $L=xl$,
$$x=\frac{(x+1)l}{lx}\iff x^2=x+1.$$
The norm of an element in $\mathbf Q(\sqrt 5)$ is the product of all the conjugates of this element. If $x=a+b\sqrt 5$ ($a,b\in\mathbf Q$), we have $N(x)=a^2-5b^2$. Thus the norm of the Golden ratio is equal to $-1$.
A: How about the original construction from the Greeks?
Let $a, b, c$ be the sides of a right triangle with $c$ as hypotenuse and $b:a=2:1$.  Then:
$b^2=c^2-a^2=(c+a)(c-a)$ (the difference of squares factorization is easily proved by the geometric methods used by the Greeks).
$c-a=(c+a)-b$ (from $b=2a$, hypothesis)
So
$b^2=(c+a)(c+a-b)$
From the cross product rule for proportions (again known since ancient times):
$(c+a)/b=b/(c+a-b)$
This represents whatcwe now call a "golden division" (a term invented later), and the ratio in the last equatyin is both $(c+a)/b$ and $b/(c-a)$.  These formulas are commonly used to geometrically construct the golden ratio, for example in Ptolemy's construction of the regular pentagon/pentagram.
