Induction on powers of powers.
$begingroup$
So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.
(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$
(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$
(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$
(Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.
Best Regards
and thanks for your time.
induction exponentiation
edited Dec 7 '18 at 19:25
Bernard
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
$endgroup$
add a comment |
$begingroup$
So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.
(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$
(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$
(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$
(Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.
Best Regards
and thanks for your time.
induction exponentiation
edited Dec 7 '18 at 19:25
Bernard
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
$endgroup$
add a comment |
$begingroup$
So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.
(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$
(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$
(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$
(Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.
Best Regards
and thanks for your time.
induction exponentiation
edited Dec 7 '18 at 19:25
Bernard
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
$endgroup$
So I would like to prove that $ a_n ; | ; a_{n+1} - 2 $, where $a_n = 6^{2^n} + 1 $ Now I know I need to do this by induction and so I begin by showing this is true for the base case.
(Proposition): P(n) := $ 6^{2^{n+1}} -1 = (6^{2^n} + 1)q ;$, where $q in mathbb{Z}$
(Base Case): P(1) is true as $ 6^{2^{2}} -1 = (6^{2} + 1)35 ;$
(Hypothesis): We assume that P(k) is True, where P(k):= $ 6^{2^{k+1}} -1 = (6^{2^k} + 1)q ;$
(Induction Step): Need to show that $P(k) implies P(k+1)$ Now I know I need to write $ 6^{2^{k+2}} -1$ in terms of my hypothesis and end up with something along the lines of $(6^{2^{k+1}} + 1)r$, where $r in mathbb{Z}$. My issue: I have and the reason I'm here is I don't feel comfortable working with powers of powers. I'm not sure how to use my hypothesis. In short looking for some guidance to get me started.
Best Regards
and thanks for your time.
induction exponentiation
induction exponentiation
edited Dec 7 '18 at 19:25
Bernard
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
edited Dec 7 '18 at 19:25
Bernard
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
edited Dec 7 '18 at 19:25
Bernard
119k740113
edited Dec 7 '18 at 19:25
Bernard
119k740113
edited Dec 7 '18 at 19:25
Bernard
119k740113
119k740113
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
asked Dec 7 '18 at 19:11
Dead_Ling0Dead_Ling0
476
476
add a comment |
add a comment |
1 Answer
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$begingroup$
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
add a comment |
$begingroup$
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
$endgroup$
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
add a comment |
$begingroup$
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
$endgroup$
You're on the right track. The key thing to use here is the difference of squares formula. In particular,
$$6^{2^{k + 1}} - 1 = (6^{2^k})^2 - 1 = (6^{2^k} - 1)(6^{2^k} + 1)$$
from which your result should follow easily.
However, induction isn't really necessary here... this gives a direct proof of divisibility.
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
answered Dec 7 '18 at 19:21
T. BongersT. Bongers
23.1k54662
23.1k54662
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
add a comment |
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
$begingroup$
Ah can't believe i didn't spot this thank you for the clarity kicking myself here !!!!
$endgroup$
– Dead_Ling0
Dec 7 '18 at 20:10
add a comment |
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