Show that number of units of $2^{2^n}+1$ is always $7$

Question

Given that $n$ is a natural number such that $n\geq2$, show that number of units of $x=2^{2^n}+1$ is always $7$

For $n=2$ we have $x=17$. For $n=3$ we have $x=257$

We can show that the number of units of $2^{2^n}$ is always $6$ too!

I don't know how can I start my proof! I don't know how can I start my proof !

Answer 1

The last digit of $2^{2^2}$ is $6$. And if $n\ge 3$ then $$\frac{2^{2^n}}{2^{2^2}}=2^{2^n-4}=2^{4k}=16^k$$ That is, $$2^{2^n}=2^{2^2}\cdot 16^k$$ Every power of a number that ends with $6$ also ends with $6$, so $2^{2^n}$ can be expressed as the product of two numbers that end with $6$. Thus, $2^{2^n}$ ends with $6$.

Answer 2

Modulo $5$: $2^{2^n}=16^{2^{n-2}}\equiv1^{2^{n-2}}=1$;
Modulo $2$: $2^{2^n}\equiv0$
so by the Chinese remainder theorem/common sense we have $2^{2^n}\equiv6\pmod{10}$. Hence $2^{2^n}+1\equiv7\pmod{10}$.

Answer 3

First, show that this is true for $n=2$:

$2^{2^2}+1=17$

Second, assume that this is true for $n$:

$2^{2^n}+1=10k+7$

Third, prove that this is true for $n+1$:

$2^{2^{n+1}}+1=$

$(\color\red{2^{2^n}+1}-1)^2+1=$

$(\color\red{10k+7}-1)^2+1=$

$100k^2+120k+37=$

$10(10k^2+12k+3)+7$

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Please note that the assumption is used only in the part marked red.