Question

Why do put at $k$ only values from $0$ to $6$

score 76 · Answer 1 · 2014-03-06 16:38:39

76

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Why do put at $k$ only values from $0$ to $6$

Published on 06 Mar 2014 - 16:38

score 55 · Answer 2 · 2026-03-30 02:07:56

55

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Proof that there are at the most two numbers of exactly six digits that squared end with the same six digits

Published on 30 Mar 2026 - 2:07

score 177 · Answer 3 · 2014-03-07 00:01:06

177

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Prove that $a \equiv b \pmod{m_1m_2}\implies a \equiv b \pmod {m_1}$

Published on 07 Mar 2014 - 0:01

score 2.2k · Answer 4 · 2014-03-07 01:26:21

2.2k

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Let $a$ and $b$ be positive integers and let $p$ be a prime number. Prove that if $a^p \equiv$ $b^p$ (mod $p$), then $a \equiv b$ (mod $p$).

Published on 07 Mar 2014 - 1:26

score 164 · Answer 5 · 2014-03-11 05:05:09

164

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Finding the remainder of a linear congruence

Published on 11 Mar 2014 - 5:05

score 181 · Answer 6 · 2014-03-11 06:58:38

181

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How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?

Published on 11 Mar 2014 - 6:58

score 56 · Answer 7 · 2026-03-26 06:19:02

56

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Congruency within the Nottingham Group

Published on 26 Mar 2026 - 6:19

score 62 · Answer 8 · 2026-03-26 06:19:58

62

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Further Congruency within the Nottingham Group

Published on 26 Mar 2026 - 6:19

score 282 · Answer 9 · 2026-03-25 17:17:28

282

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congruence mod m application

Published on 25 Mar 2026 - 17:17

score 29 · Answer 10 · 2017-04-13 12:03:28

29

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Show that number $n^2$ after division by $8$ can have remains only $0, 1$ or $4$

Published on 13 Apr 2017 - 12:03

score 52 · Answer 11 · 2017-04-13 14:17:28

52

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Show that $(a+b)^n ≡ a^n + b^n (mod 2)$ for all $a,b$ and for $n≥1$

Published on 13 Apr 2017 - 14:17

score 42 · Answer 12 · 2026-03-25 17:25:55

42

Views

Prove that $mc = mb + 8k \implies c=b+8k$ in congruence

Published on 25 Mar 2026 - 17:25

score 172 · Answer 13 · 2017-04-16 17:36:26

172

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If $p$ is prime then $(p-1)! \equiv p-1\pmod{\frac{p(p-1)}{2}} $

Published on 16 Apr 2017 - 17:36

score 2.4k · Answer 14 · 2017-04-16 18:16:39

2.4k

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$p$ prime and $d | (p-1)$ then the congruence $x^{d} \equiv 1 \pmod{p}$ has exactly $d$ incongruent solutions

Published on 16 Apr 2017 - 18:16

score 56 · Answer 15 · 2017-04-18 13:58:37

56

Views

Proof regarding Digital Roots

Published on 18 Apr 2017 - 13:58

List Question

Why do put at $k$ only values from $0$ to $6$

Proof that there are at the most two numbers of exactly six digits that squared end with the same six digits

Prove that $a \equiv b \pmod{m_1m_2}\implies a \equiv b \pmod {m_1}$

Let $a$ and $b$ be positive integers and let $p$ be a prime number. Prove that if $a^p \equiv$ $b^p$ (mod $p$), then $a \equiv b$ (mod $p$).

Finding the remainder of a linear congruence

How do $x^{p}+y^{p}=z^{p}$ and $x\equiv y \pmod{p}$ together imply $x\equiv -z \pmod{p}$?

Congruency within the Nottingham Group

Further Congruency within the Nottingham Group

congruence mod m application

Show that number $n^2$ after division by $8$ can have remains only $0, 1$ or $4$

Show that $(a+b)^n ≡ a^n + b^n (mod 2)$ for all $a,b$ and for $n≥1$

Prove that $mc = mb + 8k \implies c=b+8k$ in congruence

If $p$ is prime then $(p-1)! \equiv p-1\pmod{\frac{p(p-1)}{2}} $

$p$ prime and $d | (p-1)$ then the congruence $x^{d} \equiv 1 \pmod{p}$ has exactly $d$ incongruent solutions

Proof regarding Digital Roots

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