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While solving a binomial problem I came across this identity: $$ (a + P \sqrt{t})^n \times (a - P \sqrt{t})^n = 1 $$

where $a,P,t$ and $n$ are all integers. I pluged in some values and this identity holds, do we have a formal name or proof for this identity?

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2 Answers 2

up vote 8 down vote accepted

$$(a + P \sqrt{t})^n \times (a - P \sqrt{t})^n = (a^2 - P^2 t)^n$$

(5^2-3^2*2)^2 is not 1
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Aha thanks,I was just being a fool. –  Quixotic Apr 11 '11 at 8:34

For some values of $a, P$ and $t$ it is certainly going to work. The only thing that comes to mind if you want to give that a name is the norm in a quadratic number field. In a quadratic number field $\mathbb{Q}(\sqrt{t})$ for some square free integer $t$, every element is of the form $a + P\sqrt{t}$ with $a, P \in \mathbb{Q}$.

There's something called the norm, which is defined for any element of $\mathbb{Q}(\sqrt{t})$ by

$$\mathrm{\textbf{N}}(a + P\sqrt{t}) = (a + P\sqrt{t})(a - P\sqrt{t}) = a^2 - tP^2$$

Now, you can look at the elements $a + P\sqrt{t}$ with $a, P \in \mathbb{Z}$, and in this case the norm gives you an integer value. Well it turns out that whenever you find an element $a + P\sqrt{t}$ with $\mathrm{\textbf{N}}(a + P\sqrt{t}) = 1$, then this element is what is called a unit and basically what you have there would be $(\mathrm{\textbf{N}}(a + P\sqrt{t}) )^n = 1$.

For example in $\mathbb{Z}(\sqrt{2})$ you have $(-1 + \sqrt{2})(1 + \sqrt{2}) = 1$ and in fact $(-1 + \sqrt{2})^n (1 + \sqrt{2})^n = 1$ for any $n \in \mathbb{Z}$ and this only reflects the fact that $1 + \sqrt{2}$ is a unit in $\mathbb{Z}[\sqrt{2}]$. Maybe you should also take a look at Pell's equation which is certainly related.

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Is it necessary that t is not equal to 1 mod 4? –  quanta Apr 11 '11 at 9:22
    
Well if you want to say that $\mathbb{Z}[\sqrt{t}]$ is the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{t})$, then you certainly need to have $t \neq 1 \pmod{4}$. –  Adrián Barquero Apr 11 '11 at 9:27

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