Simplify the expression $(a+1)(a^2+1)(a^4+1)\cdots(a^{32}+1)$

Question

How do I simplify this expression?

$$\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$$

Answer 1

Oh this is so cute!

We know that $(x + y)(x - y) = x^2 - y^2$ and therefore for for any $(a^n - 1)(a^n + 1) = (a^{2n} - 1)$.

So if we simply multiply $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)$ by $(a-1)$ we get

$(a -1)\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$(a^2 - 1)\left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

And the whole thing collapses like dominoes:

$\left( a^{4}-1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$\left( a^{8}-1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$ \left( a^{16}-1 \right)\left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$ \left( a^{32}-1 \right) \left( a^{32}+1 \right)=$

$ \left( a^{64}-1 \right)$

So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right) = \frac {a^{64} - 1}{a - 1}$.

But here's are two other another interesting observations:

$(a - 1)(a^n + a^{n-1} + ...+a + 1) = (a^{n+1} + a^n + ... + a^2 + a) - (a^n + a^{n-1} +... + a + 1) = (a^{n+1} - 1)$

So $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right) = \frac {a^{64} - 1}{a - 1} = (a^{63} + a^{62} + .... + a^2 + a + 1)$

Which we can verify directly by noting

$(a^n + 1)( a^{n-1} + ...+ a + 1) = a^{2n - 1} + a^{2n - 2} + ... + a + 1$

so $\left( a+1 \right) \left( a^{2}+1 \right) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$(a^3 + a^2 + a + 1) \left( a^{4}+1 \right) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

$(a^7 + a^6+.. + a + 1) \left( a^{8}+1 \right) \left( a^{16}+1 \right) \left( a^{32}+1 \right)=$

...

$(a^{63} + a^{62}+.. + a + 1) $

Read up on proof by induction and on geometric series. They're fun stuff.

Answer 2

$$=\sum_{k=0}^{2m-1}a^{k}=\sum_{k=0}^{63}a^{k}$$ for $a\neq 1$ (geometric series) see https://en.wikipedia.org/wiki/Geometric_series $$\sum_{k=0}^{2m-1}a^{k}=\frac{1-a^{2m}}{1-a}$$

Answer 3

Hint:

Multiply and divide by $a-1$, and prove by an easy induction that $$\prod_{k=0}^n\bigl(a^{2^k}+1\bigr)=\frac{a^{2^{n+1}}-1}{a-1}.$$