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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.

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3

1

$begingroup$

Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.

• If $n$ is odd

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv -1 mod 8$

$7^n +1 equiv 0 mod 8$

Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.

• If $n$ is even

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv 1 mod 8$

$7^n +1 equiv 2 mod 8$

Therefore, the remainder of the division of $7^n+1$ is $2$.

Is that true, please?

number-theory

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asked yesterday

DimaDima

822416

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• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect {}{}{}{}{}{}{}{}{
  $endgroup$
  – hamam_Abdallah
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hamam_Abdallah Thank you so much.
  $endgroup$
  – Dima
  yesterday
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.

• If $n$ is odd

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv -1 mod 8$

$7^n +1 equiv 0 mod 8$

Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.

• If $n$ is even

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv 1 mod 8$

$7^n +1 equiv 2 mod 8$

Therefore, the remainder of the division of $7^n+1$ is $2$.

Is that true, please?

number-theory

share|cite|improve this question

asked yesterday

DimaDima

822416

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect {}{}{}{}{}{}{}{}{
  $endgroup$
  – hamam_Abdallah
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hamam_Abdallah Thank you so much.
  $endgroup$
  – Dima
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.

• If $n$ is odd

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv -1 mod 8$

$7^n +1 equiv 0 mod 8$

Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.

• If $n$ is even

$7 equiv -1 mod 8$

$7^n equiv (-1)^n mod 8$

$7^n equiv 1 mod 8$

$7^n +1 equiv 2 mod 8$

Therefore, the remainder of the division of $7^n+1$ is $2$.

Is that true, please?

number-theory

share|cite|improve this question

asked yesterday

DimaDima

822416

$endgroup$

share|cite|improve this question

asked yesterday

DimaDima

822416

share|cite|improve this question

asked yesterday

DimaDima

822416

asked yesterday

DimaDima

822416

asked yesterday

DimaDima

822416

3 Answers
3

active

oldest

votes

1

$begingroup$

Yes, you are right. Another way to present it would be like so:

For $n$ odd,
$$7^n+1 equiv (-1)^n+1 equiv -1+1 equiv 0 pmod 8$$
and for $n$ even,
$$7^n+1 equiv (-1)^n+1 equiv 1+1 equiv 2 pmod 8$$

share|cite|improve this answer

edited yesterday

answered yesterday

ThéophileThéophile

20.2k13047

$endgroup$

add a comment | 

0

$begingroup$

Note that

$forall n in Bbb N, ; 7^n + 1 = (8 - 1)^n + 1 = displaystyle sum_0^n dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , tag 1$

via the binomial theorem; when $n$ is odd this yields

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k , tag 2$

whence

$8 mid 7^n + 1, ; text{odd} ; n; tag 3$

for even $n$ we obtain

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k + 2; tag 4$

we note that

$0 le 2 < 8; tag 5$

that is, the remainder of $7^n + 1$ when divided by $8$ is $2$.

share|cite|improve this answer

edited yesterday

answered yesterday

Robert LewisRobert Lewis

47.8k23067

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure why anyone would think that is a better way than what the OP did which was easy, clear and correct.
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Just to be different:

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

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

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

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

So we $a = 7$ we get $7+1=8|a^n+1$ if $n$ is odd. And $7+1=8|a^n-1$ if $n$ is even. And as $a^n-1 = (a^n -1) + 2$, $a^n + 1$ will have remainder $2$ if $n$ is even.

But that's just to be different.

What you did was shorter and better. This just shows why it's no surprise.

share|cite|improve this answer

answered yesterday

fleabloodfleablood

72k22687

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add a comment | 

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1

$begingroup$

Yes, you are right. Another way to present it would be like so:

For $n$ odd,
$$7^n+1 equiv (-1)^n+1 equiv -1+1 equiv 0 pmod 8$$
and for $n$ even,
$$7^n+1 equiv (-1)^n+1 equiv 1+1 equiv 2 pmod 8$$

share|cite|improve this answer

edited yesterday

answered yesterday

ThéophileThéophile

20.2k13047

$endgroup$

add a comment | 

1

$begingroup$

Yes, you are right. Another way to present it would be like so:

For $n$ odd,
$$7^n+1 equiv (-1)^n+1 equiv -1+1 equiv 0 pmod 8$$
and for $n$ even,
$$7^n+1 equiv (-1)^n+1 equiv 1+1 equiv 2 pmod 8$$

share|cite|improve this answer

edited yesterday

answered yesterday

ThéophileThéophile

20.2k13047

$endgroup$

add a comment | 

$begingroup$

Yes, you are right. Another way to present it would be like so:

For $n$ odd,
$$7^n+1 equiv (-1)^n+1 equiv -1+1 equiv 0 pmod 8$$
and for $n$ even,
$$7^n+1 equiv (-1)^n+1 equiv 1+1 equiv 2 pmod 8$$

share|cite|improve this answer

edited yesterday

answered yesterday

ThéophileThéophile

20.2k13047

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

ThéophileThéophile

20.2k13047

answered yesterday

ThéophileThéophile

20.2k13047

answered yesterday

ThéophileThéophile

20.2k13047

0

$begingroup$

Note that

$forall n in Bbb N, ; 7^n + 1 = (8 - 1)^n + 1 = displaystyle sum_0^n dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , tag 1$

via the binomial theorem; when $n$ is odd this yields

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k , tag 2$

whence

$8 mid 7^n + 1, ; text{odd} ; n; tag 3$

for even $n$ we obtain

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k + 2; tag 4$

we note that

$0 le 2 < 8; tag 5$

that is, the remainder of $7^n + 1$ when divided by $8$ is $2$.

share|cite|improve this answer

edited yesterday

answered yesterday

Robert LewisRobert Lewis

47.8k23067

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure why anyone would think that is a better way than what the OP did which was easy, clear and correct.
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Note that

$forall n in Bbb N, ; 7^n + 1 = (8 - 1)^n + 1 = displaystyle sum_0^n dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , tag 1$

via the binomial theorem; when $n$ is odd this yields

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k , tag 2$

whence

$8 mid 7^n + 1, ; text{odd} ; n; tag 3$

for even $n$ we obtain

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k + 2; tag 4$

we note that

$0 le 2 < 8; tag 5$

that is, the remainder of $7^n + 1$ when divided by $8$ is $2$.

share|cite|improve this answer

edited yesterday

answered yesterday

Robert LewisRobert Lewis

47.8k23067

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Not sure why anyone would think that is a better way than what the OP did which was easy, clear and correct.
  $endgroup$
  – fleablood
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Note that

$forall n in Bbb N, ; 7^n + 1 = (8 - 1)^n + 1 = displaystyle sum_0^n dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + 1 = sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k}(-1)^k + (-1)^n + 1 , tag 1$

via the binomial theorem; when $n$ is odd this yields

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k , tag 2$

whence

$8 mid 7^n + 1, ; text{odd} ; n; tag 3$

for even $n$ we obtain

$7^n + 1 = displaystyle 8sum_0^{n - 1} dfrac{n!}{k!(n - k)!}8^{n - k - 1}(-1)^k + 2; tag 4$

we note that

$0 le 2 < 8; tag 5$

that is, the remainder of $7^n + 1$ when divided by $8$ is $2$.

share|cite|improve this answer

edited yesterday

answered yesterday

Robert LewisRobert Lewis

47.8k23067

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Robert LewisRobert Lewis

47.8k23067

answered yesterday

Robert LewisRobert Lewis

47.8k23067

answered yesterday

Robert LewisRobert Lewis

47.8k23067

0

$begingroup$

Just to be different:

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

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

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

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

So we $a = 7$ we get $7+1=8|a^n+1$ if $n$ is odd. And $7+1=8|a^n-1$ if $n$ is even. And as $a^n-1 = (a^n -1) + 2$, $a^n + 1$ will have remainder $2$ if $n$ is even.

But that's just to be different.

What you did was shorter and better. This just shows why it's no surprise.

share|cite|improve this answer

answered yesterday

fleabloodfleablood

72k22687

$endgroup$

add a comment | 

0

$begingroup$

Just to be different:

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

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

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

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

So we $a = 7$ we get $7+1=8|a^n+1$ if $n$ is odd. And $7+1=8|a^n-1$ if $n$ is even. And as $a^n-1 = (a^n -1) + 2$, $a^n + 1$ will have remainder $2$ if $n$ is even.

But that's just to be different.

What you did was shorter and better. This just shows why it's no surprise.

share|cite|improve this answer

answered yesterday

fleabloodfleablood

72k22687

$endgroup$

add a comment | 

$begingroup$

Just to be different:

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

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

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

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

So we $a = 7$ we get $7+1=8|a^n+1$ if $n$ is odd. And $7+1=8|a^n-1$ if $n$ is even. And as $a^n-1 = (a^n -1) + 2$, $a^n + 1$ will have remainder $2$ if $n$ is even.

But that's just to be different.

What you did was shorter and better. This just shows why it's no surprise.

share|cite|improve this answer

answered yesterday

fleabloodfleablood

72k22687

$endgroup$

share|cite|improve this answer

answered yesterday

fleabloodfleablood

72k22687

answered yesterday

fleabloodfleablood

72k22687

answered yesterday

fleabloodfleablood

72k22687