Find $7^{2018}-1 bmod 5$ using Fermat's Little Theorem
$begingroup$
Let $z=7^{2018}-1$
Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$
$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?
Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.
elementary-number-theory modular-arithmetic
$endgroup$
add a comment |
$begingroup$
Let $z=7^{2018}-1$
Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$
$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?
Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.
elementary-number-theory modular-arithmetic
$endgroup$
2
$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
1
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49
add a comment |
$begingroup$
Let $z=7^{2018}-1$
Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$
$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?
Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.
elementary-number-theory modular-arithmetic
$endgroup$
Let $z=7^{2018}-1$
Theorem. : Let $p in mathbb{N}$ be any prime and $a in mathbb{Z}$. If $p$ doesn't divide $a$, then $p$ divides $a^{p-1}-1$
$$
z=left(7^{1009}right)^2-1=left(7^{1009}right)^{3-1}-1
$$
For $p=3$ and $a=7^{1009}$ the theorem gives:
$$
z bmod 3=0
$$
Any hints on how to proceed to find $z bmod 5$?
Note: Some kind of error occurred and the important part of the last sentence got deleted/never composed. I apologize for the ambiguity of my question in its previous form.
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
edited Dec 15 '18 at 17:22
Jevaut
asked Dec 15 '18 at 13:36
JevautJevaut
1,156112
1,156112
2
$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
1
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49
add a comment |
2
$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
1
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49
2
2
$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
1
1
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Any hints on how to proceed to find zmod5?
Do it the EXACT same way.
$5 -1 = X$
And $2018div X = Q$ with $R$ remaider.
So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.
Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.
.....
But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)
$endgroup$
add a comment |
$begingroup$
We have:
$$(7^2)^{1009} - 1 pmod 5$$
$$equiv (-1)^{1009} - 1$$
$$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$
$endgroup$
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
add a comment |
$begingroup$
$7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$
$7^2=49 ≡-1mod 5$
$7^{2}.7^{2016}=7^{2018} ≡-1mod 5$
⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any hints on how to proceed to find zmod5?
Do it the EXACT same way.
$5 -1 = X$
And $2018div X = Q$ with $R$ remaider.
So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.
Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.
.....
But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)
$endgroup$
add a comment |
$begingroup$
Any hints on how to proceed to find zmod5?
Do it the EXACT same way.
$5 -1 = X$
And $2018div X = Q$ with $R$ remaider.
So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.
Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.
.....
But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)
$endgroup$
add a comment |
$begingroup$
Any hints on how to proceed to find zmod5?
Do it the EXACT same way.
$5 -1 = X$
And $2018div X = Q$ with $R$ remaider.
So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.
Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.
.....
But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)
$endgroup$
Any hints on how to proceed to find zmod5?
Do it the EXACT same way.
$5 -1 = X$
And $2018div X = Q$ with $R$ remaider.
So $7^{2018}-1 = (7^{Q})^X*7^R - 1 equiv 7^R - 1 pmod 5$.
Figuring $7^R-1$ can be done by hand as $7equiv 2pmod 5$ so $7^Requiv 2^Rpmod 7$ and $R < 5$.
.....
But I do have advice: Don't to $7^{2018} - 1 = (7^{2018})^2 - 1 equiv 1-1 equiv 0 pmod 3$. Do $7^{2018} -1 = (7^2)^{2018} - 1 equiv 1^{2018} - 1equiv 0 pmod 3$. It'll make things easier when $2018div (p-1)$ has a remainder (as it does with $p = 5$)
answered Dec 15 '18 at 17:30
fleabloodfleablood
70.5k22685
70.5k22685
add a comment |
add a comment |
$begingroup$
We have:
$$(7^2)^{1009} - 1 pmod 5$$
$$equiv (-1)^{1009} - 1$$
$$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$
$endgroup$
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
add a comment |
$begingroup$
We have:
$$(7^2)^{1009} - 1 pmod 5$$
$$equiv (-1)^{1009} - 1$$
$$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$
$endgroup$
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
add a comment |
$begingroup$
We have:
$$(7^2)^{1009} - 1 pmod 5$$
$$equiv (-1)^{1009} - 1$$
$$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$
$endgroup$
We have:
$$(7^2)^{1009} - 1 pmod 5$$
$$equiv (-1)^{1009} - 1$$
$$equiv -2 equiv bbox[5px,border:2px solid black]3 pmod 5$$
edited Dec 15 '18 at 14:06
answered Dec 15 '18 at 13:46
Toby MakToby Mak
3,49811128
3,49811128
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
add a comment |
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
@BillDubuque Thanks!
$endgroup$
– Toby Mak
Dec 15 '18 at 14:07
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
Is $a^n equiv (a bmod 5)^n , , (bmod 5)$ a general rule?
$endgroup$
– Jevaut
Dec 15 '18 at 15:21
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
$begingroup$
"Is an≡(amod5)n(mod5) a general rule?" Yes. Arithmetic (addition, subtraction, multiplication, and raising things to powers) distributes over modulo. (Although division requires care; and being powers absolutely does not). That's a very basic fact you should have proven for yourself long before learning FLT.
$endgroup$
– fleablood
Dec 15 '18 at 17:38
add a comment |
$begingroup$
$7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$
$7^2=49 ≡-1mod 5$
$7^{2}.7^{2016}=7^{2018} ≡-1mod 5$
⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$
$endgroup$
add a comment |
$begingroup$
$7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$
$7^2=49 ≡-1mod 5$
$7^{2}.7^{2016}=7^{2018} ≡-1mod 5$
⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$
$endgroup$
add a comment |
$begingroup$
$7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$
$7^2=49 ≡-1mod 5$
$7^{2}.7^{2016}=7^{2018} ≡-1mod 5$
⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$
$endgroup$
$7^{5-1}≡1mod 5$ ⇒ $7^{2016}=(7^4)^{504} ≡1mod 5$
$7^2=49 ≡-1mod 5$
$7^{2}.7^{2016}=7^{2018} ≡-1mod 5$
⇒ $2^{2018}-1≡-2mod 5 ≡3mod 5$
answered Dec 15 '18 at 17:59
siroussirous
1,6611514
1,6611514
add a comment |
add a comment |
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$begingroup$
I think you mean "if $p$ doesn't divide $a$". But why are you working $pmod 3$ when the question is asking about $pmod 5$?
$endgroup$
– lulu
Dec 15 '18 at 13:38
1
$begingroup$
Also a duplicate of this.
$endgroup$
– Jyrki Lahtonen
Dec 15 '18 at 13:49