I was reading the book Seventeen equations have changed the world.
At some point, while the book was talking about complex numbers, I see this equation:
$2+ \sqrt{-121} = (2+ \sqrt{-1})^3$
Even if it's easy to proof the truth of this equivalence (it is enough to develop the two members),
I can't find an easy/good/fast way to obtain straight the identity.
Can you help me? Does there exist a mathematical property that I'm missing?
On
Not sure how you approached it before.
If you want to utlise the binomial expansion $$ (x+y)^n = \sum_{k=0}^n\left(\matrix{n\\k}\right)x^ky^{n-k} $$ then we have $$ \begin{align} (x+y)^3 &= \left(\matrix{3\\0}\right)2^3 +\left(\matrix{3\\1}\right)2^2\sqrt{-1} + \left(\matrix{3\\2}\right)2^1(\sqrt{-1})^2 + \left(\matrix{3\\3}\right)(\sqrt{-1})^3 \\ &= 2^3 + 3\cdot 2^2\sqrt{-1} + 3\cdot 2\cdot-1 + \sqrt{-1}\cdot(\sqrt{-1})^2\\ &=8+12\sqrt{-1} -6 -\sqrt{-1} = 2 +11\sqrt{-1} = 2 + \sqrt{11^2\cdot - 1} \\ &=2+\sqrt{-121} \end{align} $$
Expanding the product as \begin{eqnarray*} (2+\sqrt{-1})^3&=&2^3+3\times2^2\sqrt{-1}+3\times2\sqrt{-1}^2+\sqrt{-1}^3\\ &=&(2^3-6)+(3\times2^2-1)\sqrt{-1}\\ &=&2+11\sqrt{-1}\\ &=&2+\sqrt{-121}, \end{eqnarray*} is so elementary that I cannot think of any easier way to show it.
But suppose you want to determine the factorization of $2+\sqrt{-121}$, i.e. roughly speaking an expression $$2+\sqrt{-121}=(a_1+b_1\sqrt{-1})(a_2+b_2\sqrt{-1})\cdots(a_n+b_n\sqrt{-1}),$$ where the $a_i$ and $b_i$ are integers. This is the subject of algebraic number theory. I will use a few results without mention, as I don't know your background in the subject. You could note that the norm equals $$N(2+\sqrt{-121})=N(2+11\sqrt{-1})=2^2+11^2=125=5^3,$$ and that $5$ does not divide $2+\sqrt{-121}$. This implies that $$2+11\sqrt{-1}=(a+b\sqrt{-1})^3,$$ where the norm of $a+b\sqrt{-1}$ equals $5$. This means that $$5=N(a+b\sqrt{-1})=a^2+b^2,$$ which has the solutions $$1+2\sqrt{-1},\qquad 1-2\sqrt{-1},\qquad 2+\sqrt{-1},\qquad 2-\sqrt{-1}.$$ It is then a quick check to show that only $2+\sqrt{-1}$ works.