Mixing
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Prove the divisibility of $R^k_npm 1$ by 4
The question is:
Prove that:
For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.
My Try:
- First we try to find $a$ such that $R_nequiv a pmod 4$.
Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:
$$a equiv -1 pmod 4$$
- The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$
Any ideas or notes on the prove?
Is there simpler way to prove so?
elementary-number-theory modular-arithmetic divisibility
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
add a comment |
The question is:
Prove that:
For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.
My Try:
- First we try to find $a$ such that $R_nequiv a pmod 4$.
Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:
$$a equiv -1 pmod 4$$
- The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$
Any ideas or notes on the prove?
Is there simpler way to prove so?
elementary-number-theory modular-arithmetic divisibility
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
1
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28
add a comment |
The question is:
Prove that:
For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.
My Try:
- First we try to find $a$ such that $R_nequiv a pmod 4$.
Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:
$$a equiv -1 pmod 4$$
- The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$
Any ideas or notes on the prove?
Is there simpler way to prove so?
elementary-number-theory modular-arithmetic divisibility
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
The question is:
Prove that:
For $n>1, R_n^k-1$ is divisible by $4$ if $k$ is even. If $k$ is odd, then $R_n^k+1$ is divisible by $4$ where $R_n$ is the rep-unit of length $n$ i.e. $R_3 = 111 , R_5 = 11111, cdots$.
My Try:
- First we try to find $a$ such that $R_nequiv a pmod 4$.
Notice that we can express these integers as $R_n = frac{10^n-1}{9}$. Now:
$$frac{10^n-1}{9} equiv a pmod 4$$
$$10^n-1 equiv 9a pmod 4 to 10^n-1 equiv a pmod 4$$
But $10^n$ is divisible by $4$ for any $n>1$. Hence:
$$0-1 equiv a pmod 4$$
Implying:
$$a equiv -1 pmod 4$$
- The second step is trivial. If $k$ is odd, then $(-1)^k +1 = -1 + 1 = 0 pmod 4$, else if $k$ is even, then $(-1)^k -1 = 1-1 = 0 pmod 4$
Any ideas or notes on the prove?
Is there simpler way to prove so?
elementary-number-theory modular-arithmetic divisibility
elementary-number-theory modular-arithmetic divisibility
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
asked Nov 22 '18 at 12:57
Maged SaeedMaged Saeed
8471417
8471417
1
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28
add a comment |
1
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28
1
1
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28
add a comment |
1 Answer
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Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies
$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus
$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$
therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
add a comment |
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1 Answer
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Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies
$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus
$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$
therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
add a comment |
Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies
$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus
$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$
therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
add a comment |
Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies
$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus
$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$
therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
Your proof is correct. As for "simpler ways", we may use modular fractions. If $,gcd(b,m)=1,$ then $,b^{-1}$ exists $!bmod m,$ and $,a/b := ab^{-1}$ is a unique solution of $,bxequiv apmod{!m},,$ and it satisfies
$!bmod m!:, begin{align}A&equiv a\ B&equiv bend{align} Rightarrow dfrac{A}Bequiv dfrac{a}b equiv dfrac{abmod m}{bbmod m}, {bf if} gcd(b,m)=1; $ thus
$bmod color{#c00}4!:, R_n := dfrac{(2i)^{large n}-1}{4j+1}, =, dfrac{color{#c00}4i^2 (2i)^{large n-2}-1}{color{#c00}4j+1},equiv, dfrac{-1}1 $ for all $,nge 2$
therefore $ R_n^{large k},equiv, (-1)^{large k} $ by Congruence Power Rule. $ $ OP has $,i = 5,, j = 2$
Beware $ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
edited Nov 22 '18 at 16:34
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
answered Nov 22 '18 at 16:18
Bill DubuqueBill Dubuque
209k29190631
209k29190631
add a comment |
add a comment |
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1
To prove $R_n equiv -1 pmod 4$, you can use the fact that $R_n = 100 * alpha + 11 $ which can save you a few lines. The second step is really straightforward and can't be improved IMHO
– F.Carette
Nov 22 '18 at 13:06
@F.Carette, Thanks.
– Maged Saeed
Nov 22 '18 at 13:07
@F.Carette It's improvable by improvng the proposition to: $ 4mid R^k - (-1)^k. $ Then $!bmod 4!:, Requiv -1,Rightarrow, R^kequiv (-1)^k,$ follows immediately by the Congruence Power Rule.
– Bill Dubuque
Nov 22 '18 at 15:28