Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that by induction method that $2^{2^n}+1$ has $7$ in unit's place for all $n\geq 2$.

I have tried to show this with the following way :

Let $f(n)=2^{2^n}+1$.
Then for $n=2,f(2)=2^{2^2}+1=17\Rightarrow f(2)\equiv 7(\mod 10) $
Suppose for $n=m$, the result is true i.e., $f(m)=2^{2^m}+1=10p+7$, where $p$ is an integer.
How can I show this result for $n=m+1$ ?

share|improve this question
    
$f(1)=5$; you want $f(2)$ in that base case calculation. –  Brian M. Scott Aug 15 '13 at 17:16
    
You’re making much too big a deal of this, I believe. Did you actually evaluate $2^{2^2}$ and $2^{2^3}$? –  Lubin Aug 15 '13 at 17:18
    
Sorry for my mistake. A correction has been done. –  Md Saiful Gazi Aug 15 '13 at 17:19

5 Answers 5

HINT: $$f(m+1)=2^{2^{m+1}}+1=2^{2^m\cdot2}+1=\left(2^{2^m}\right)^2+1=\big(f(m)-1\big)^2+1$$

share|improve this answer

Hint: Note that $2^{2^{n+1}}=\left(2^{2^n}\right)^2$ and $(10k+6)^2=100k^2+120 k+36$

share|improve this answer

HINT: When you square a number ending in $6$, don’t you get another such number?

share|improve this answer

Hint: apply induction to $2^{2^m}$

Additional possibility: investigate the units digits of small powers of $2$ and formulate and prove a stronger (but simpler) hypothesis by induction of which the given statement is a special case.

share|improve this answer

Let the statement F(n) be true and with F(n+1) the power of 2 that is 2^m becomes 2^(n+1) which is a number of the form 4k. And 2^4k will always yield a 6 in its last digit and by adding 1 you will get 7 in the last digit.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.