Let a be any odd positive integer

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Let a be any odd positive integer and n be an integer greater than 5 .What is the smallest possible integer N such that $$a^N $$ is congruent to 1 modulo 2^n




number-theory




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  • What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
    – Toby Mak
    Jul 20 at 7:08










  • I tried it with eulers phi function but couldn't proceed
    – Shrimon Mukherjee
    Jul 20 at 7:08










  • How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
    – Toby Mak
    Jul 20 at 7:12















up vote
0
down vote

favorite












Let a be any odd positive integer and n be an integer greater than 5 .What is the smallest possible integer N such that $$a^N $$ is congruent to 1 modulo 2^n




number-theory




share|cite|improve this question



















  • What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
    – Toby Mak
    Jul 20 at 7:08










  • I tried it with eulers phi function but couldn't proceed
    – Shrimon Mukherjee
    Jul 20 at 7:08










  • How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
    – Toby Mak
    Jul 20 at 7:12













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let a be any odd positive integer and n be an integer greater than 5 .What is the smallest possible integer N such that $$a^N $$ is congruent to 1 modulo 2^n




number-theory




share|cite|improve this question











Let a be any odd positive integer and n be an integer greater than 5 .What is the smallest possible integer N such that $$a^N $$ is congruent to 1 modulo 2^n





number-theory





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 20 at 6:57









Shrimon Mukherjee

175




175











  • What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
    – Toby Mak
    Jul 20 at 7:08










  • I tried it with eulers phi function but couldn't proceed
    – Shrimon Mukherjee
    Jul 20 at 7:08










  • How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
    – Toby Mak
    Jul 20 at 7:12

















  • What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
    – Toby Mak
    Jul 20 at 7:08










  • I tried it with eulers phi function but couldn't proceed
    – Shrimon Mukherjee
    Jul 20 at 7:08










  • How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
    – Toby Mak
    Jul 20 at 7:12
















What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
– Toby Mak
Jul 20 at 7:08




What happens if $a = 1$? As this problem is related to modular arithmetic, have you tried applying any properties and rules?
– Toby Mak
Jul 20 at 7:08












I tried it with eulers phi function but couldn't proceed
– Shrimon Mukherjee
Jul 20 at 7:08




I tried it with eulers phi function but couldn't proceed
– Shrimon Mukherjee
Jul 20 at 7:08












How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
– Toby Mak
Jul 20 at 7:12





How about $1^6 pmod 2^6$? If not, how about $(1+2^6)^6 pmod 2^6$?
– Toby Mak
Jul 20 at 7:12











1 Answer
1






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up vote
1
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Hint: This is $mathrmord_2^n(a)$. We know that



$$mathrmord_2^n(a)|varphi(2^n)=2^n-1,$$



so this number must be a power of $2$. Now see if you can apply lifting the exponent lemma to this to find the number of times $2$ divides $a^2^k-1$ for a given $k$.






share|cite|improve this answer





















  • Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
    – lhf
    Jul 20 at 10:49











  • @lhf This is true, but not necessary for this solution.
    – Carl Schildkraut
    Jul 20 at 18:09










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1 Answer
1






active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Hint: This is $mathrmord_2^n(a)$. We know that



$$mathrmord_2^n(a)|varphi(2^n)=2^n-1,$$



so this number must be a power of $2$. Now see if you can apply lifting the exponent lemma to this to find the number of times $2$ divides $a^2^k-1$ for a given $k$.






share|cite|improve this answer





















  • Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
    – lhf
    Jul 20 at 10:49











  • @lhf This is true, but not necessary for this solution.
    – Carl Schildkraut
    Jul 20 at 18:09














up vote
1
down vote













Hint: This is $mathrmord_2^n(a)$. We know that



$$mathrmord_2^n(a)|varphi(2^n)=2^n-1,$$



so this number must be a power of $2$. Now see if you can apply lifting the exponent lemma to this to find the number of times $2$ divides $a^2^k-1$ for a given $k$.






share|cite|improve this answer





















  • Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
    – lhf
    Jul 20 at 10:49











  • @lhf This is true, but not necessary for this solution.
    – Carl Schildkraut
    Jul 20 at 18:09












up vote
1
down vote










up vote
1
down vote









Hint: This is $mathrmord_2^n(a)$. We know that



$$mathrmord_2^n(a)|varphi(2^n)=2^n-1,$$



so this number must be a power of $2$. Now see if you can apply lifting the exponent lemma to this to find the number of times $2$ divides $a^2^k-1$ for a given $k$.






share|cite|improve this answer













Hint: This is $mathrmord_2^n(a)$. We know that



$$mathrmord_2^n(a)|varphi(2^n)=2^n-1,$$



so this number must be a power of $2$. Now see if you can apply lifting the exponent lemma to this to find the number of times $2$ divides $a^2^k-1$ for a given $k$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 20 at 7:12









Carl Schildkraut

8,26211238




8,26211238











  • Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
    – lhf
    Jul 20 at 10:49











  • @lhf This is true, but not necessary for this solution.
    – Carl Schildkraut
    Jul 20 at 18:09
















  • Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
    – lhf
    Jul 20 at 10:49











  • @lhf This is true, but not necessary for this solution.
    – Carl Schildkraut
    Jul 20 at 18:09















Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
– lhf
Jul 20 at 10:49





Actually, for $nge 3$, we have $mathrmord_2^n(a)midlambda(2^n)=2^n-2$.
– lhf
Jul 20 at 10:49













@lhf This is true, but not necessary for this solution.
– Carl Schildkraut
Jul 20 at 18:09




@lhf This is true, but not necessary for this solution.
– Carl Schildkraut
Jul 20 at 18:09












 

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