How to reduce the congruence $(p-1)^{p-1}equiv 2^{p-1}pmod{p^2}$ into $p(p-1)+2^{p-1}-1equiv 0pmod{p^2}$?












0












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In addition, how is $(p-a)^{p-1}-a^{p-1}equiv -(p-1)pa^{p-2}pmod{p^2}$ derived by binomial theorem?






modular-arithmetic







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  • 1




    $begingroup$
    What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
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    – postmortes
    Jan 17 at 19:26
















0












$begingroup$


In addition, how is $(p-a)^{p-1}-a^{p-1}equiv -(p-1)pa^{p-2}pmod{p^2}$ derived by binomial theorem?






modular-arithmetic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
    $endgroup$
    – postmortes
    Jan 17 at 19:26














0












0








0


0



$begingroup$


In addition, how is $(p-a)^{p-1}-a^{p-1}equiv -(p-1)pa^{p-2}pmod{p^2}$ derived by binomial theorem?






modular-arithmetic







share|cite|improve this question











$endgroup$




In addition, how is $(p-a)^{p-1}-a^{p-1}equiv -(p-1)pa^{p-2}pmod{p^2}$ derived by binomial theorem?






modular-arithmetic




modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 19:08









Bernard

121k740116




121k740116










asked Jan 17 at 18:58









TriHard 7TriHard 7

1




1








  • 1




    $begingroup$
    What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
    $endgroup$
    – postmortes
    Jan 17 at 19:26














  • 1




    $begingroup$
    What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
    $endgroup$
    – postmortes
    Jan 17 at 19:26








1




1




$begingroup$
What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
$endgroup$
– postmortes
Jan 17 at 19:26




$begingroup$
What have you tried so far? Questions that lack context and some description of what you've tried and where you got stuck tend to attract downvotes.
$endgroup$
– postmortes
Jan 17 at 19:26










2 Answers
2






active

oldest

votes


















1












$begingroup$

Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 equiv 2^{p-1}$, so $0 equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Didn't think it would be this straightforward.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:45



















0












$begingroup$

Note that, by the binomial theorem
$$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+binom{p-1}2p^2-cdots$$
(at least if $p$ is odd).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks, the original problem indeed stated that p is a prime greater than 2.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:43











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 equiv 2^{p-1}$, so $0 equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Didn't think it would be this straightforward.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:45
















1












$begingroup$

Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 equiv 2^{p-1}$, so $0 equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Didn't think it would be this straightforward.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:45














1












1








1





$begingroup$

Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 equiv 2^{p-1}$, so $0 equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.






share|cite|improve this answer









$endgroup$



Note that $(p-1)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2} + ... - (p-1)p^1 + 1$, and only the last two cannot be divided by $p^2$ , so you are left with $-(p-1)p^1 + 1 equiv 2^{p-1}$, so $0 equiv 2^{p-1} +p(p-1) - 1$. As for the second thing, you have $(p-a)^{p-1}$ = $p^{p-1} - (p-1)p^{p-2}a^1 + ... -(p-1)p^1a^{p-2} + a^p$, and you can proceed as before.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 19:32









Jakub AndruszkiewiczJakub Andruszkiewicz

2116




2116












  • $begingroup$
    Thanks! Didn't think it would be this straightforward.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:45


















  • $begingroup$
    Thanks! Didn't think it would be this straightforward.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:45
















$begingroup$
Thanks! Didn't think it would be this straightforward.
$endgroup$
– TriHard 7
Jan 17 at 19:45




$begingroup$
Thanks! Didn't think it would be this straightforward.
$endgroup$
– TriHard 7
Jan 17 at 19:45











0












$begingroup$

Note that, by the binomial theorem
$$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+binom{p-1}2p^2-cdots$$
(at least if $p$ is odd).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks, the original problem indeed stated that p is a prime greater than 2.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:43
















0












$begingroup$

Note that, by the binomial theorem
$$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+binom{p-1}2p^2-cdots$$
(at least if $p$ is odd).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks, the original problem indeed stated that p is a prime greater than 2.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:43














0












0








0





$begingroup$

Note that, by the binomial theorem
$$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+binom{p-1}2p^2-cdots$$
(at least if $p$ is odd).






share|cite|improve this answer









$endgroup$



Note that, by the binomial theorem
$$(p-1)^{p-1}=(1-p)^{p-1}=1-(p-1)p+binom{p-1}2p^2-cdots$$
(at least if $p$ is odd).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 19:31









Lord Shark the UnknownLord Shark the Unknown

105k1160132




105k1160132












  • $begingroup$
    thanks, the original problem indeed stated that p is a prime greater than 2.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:43


















  • $begingroup$
    thanks, the original problem indeed stated that p is a prime greater than 2.
    $endgroup$
    – TriHard 7
    Jan 17 at 19:43
















$begingroup$
thanks, the original problem indeed stated that p is a prime greater than 2.
$endgroup$
– TriHard 7
Jan 17 at 19:43




$begingroup$
thanks, the original problem indeed stated that p is a prime greater than 2.
$endgroup$
– TriHard 7
Jan 17 at 19:43


















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