Denominator of Wolstenholme's Theorem [closed]Show that the numerator of $sum_{k=1}^{p^2} frac1k -...

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Denominator of Wolstenholme's Theorem [closed]


Show that the numerator of $sum_{k=1}^{p^2} frac1k - frac1psum_{k=1}^pfrac1k$ is divisible by $p^4$A proof of Wolstenholme's theoremProof of Wolstenholme's theoremLucas' Theorem for $p$-adic integers?Chinese remainder theorem other way aroundTo prove ${2p - 1 choose p } equiv 1 pmod{p^2}$ without using Wolstenholme's theoremLet $p>2$ be a prime, $f(x)=sum_{k=1}^{p-1}x^k/k$, $f(2)-f(-1)=m/n$, where $gcd(m,n)=1$. Show that $pmid m$.Bernoulli numbers : a strengthening of von Staudt's theorem.facing a small obstacle while trying to proof Wolstenholme's theorem 1 using multiplicative inverse in (mod p)Special case of Wolstenholme's theorem true for p=3?













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$begingroup$


For irreducible fraction $ frac{A}{B} =sum_{k=1}^{p-1} frac{1}{k} $ ,



$A equiv 0 (mod p^2 )$
by Wolstenholme's Theorem,



Then, how can I find out $B (mod p^2 )$ ?










share|cite|improve this question











$endgroup$



closed as off-topic by Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos Mar 13 at 8:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
    $endgroup$
    – b00n heT
    Mar 13 at 6:49


















-1












$begingroup$


For irreducible fraction $ frac{A}{B} =sum_{k=1}^{p-1} frac{1}{k} $ ,



$A equiv 0 (mod p^2 )$
by Wolstenholme's Theorem,



Then, how can I find out $B (mod p^2 )$ ?










share|cite|improve this question











$endgroup$



closed as off-topic by Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos Mar 13 at 8:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
    $endgroup$
    – b00n heT
    Mar 13 at 6:49
















-1












-1








-1





$begingroup$


For irreducible fraction $ frac{A}{B} =sum_{k=1}^{p-1} frac{1}{k} $ ,



$A equiv 0 (mod p^2 )$
by Wolstenholme's Theorem,



Then, how can I find out $B (mod p^2 )$ ?










share|cite|improve this question











$endgroup$




For irreducible fraction $ frac{A}{B} =sum_{k=1}^{p-1} frac{1}{k} $ ,



$A equiv 0 (mod p^2 )$
by Wolstenholme's Theorem,



Then, how can I find out $B (mod p^2 )$ ?







number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 13 at 18:44







bthrough

















asked Mar 13 at 6:39









bthroughbthrough

134




134




closed as off-topic by Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos Mar 13 at 8:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos Mar 13 at 8:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Saad, John Omielan, Cesareo, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
    $endgroup$
    – b00n heT
    Mar 13 at 6:49




















  • $begingroup$
    Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
    $endgroup$
    – b00n heT
    Mar 13 at 6:49


















$begingroup$
Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
$endgroup$
– b00n heT
Mar 13 at 6:49






$begingroup$
Hint: From Wikipedia's page >As Wolstenholme himself established, his theorem can also be expressed as a pair of congruences for (generalized) harmonic numbers: $$1+{1 over 2}+{1 over 3}+...+{1 over p-1} equiv 0 pmod{p^2} $$ This looks promising, but I am not confident enough with the subject to claim more.
$endgroup$
– b00n heT
Mar 13 at 6:49












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