Prove that any number consisting of $2^{n}$ identical digits has at least $n$ distinct prime factors.

Question

Prove that any number consisting of $2^{n}$ identical digits has at least $n$ distinct prime factors.

Proof: Such a number $N$ can be written as $ N=k \cdot \frac{10^{2^{n}}-1}{10-1}=k(10+1)\left(10^{2}+1\right) \cdots\left(10^{2^{n-1}}+1\right) $

but how they write $k \cdot \frac{10^{2^{n}}-1}{10-1}=k(10+1)\left(10^{2}+1\right) \cdots\left(10^{2^{n-1}}+1\right) $

I think i am missing some identity , can someone pls tell me what is it ?

Answer 1

It's repeated use of the difference-of-squares identity $(x-1)(x+1) = x^2-1$.

We have $(10-1)(10+1) = 10^2-1$.

Therefore $(10-1)(10+1)(10^2+1) = (10^2-1)(10^2+1) = 10^4-1$.

Therefore $(10-1)(10+1)(10^2+1)(10^4+1) = (10^4-1)(10^4+1) = 10^8-1$.

Repeating this, $(10-1)(10+1)(10^2+1)(\dotsm)(10^{2^{n-1}}+1) = 10^{2^n} - 1$.

Finally, divide both sides by $10-1$, and we get a factorization of $\frac{10^{2^n} - 1}{10-1}$ into $n$ factors: $10^{2^0}+1$ through $10^{2^{n-1}} + 1$.