Question

How to solve the below system of congruence?

score 49 · Answer 1 · 2018-04-02 22:38:35

49

Views

How to solve the below system of congruence?

Published on 02 Apr 2018 - 22:38

score 48 · Answer 2 · 2018-04-03 09:35:39

48

Views

Odd integer is power of 3 modulo power of 2

Published on 03 Apr 2018 - 9:35

score 26 · Answer 3 · 2018-04-03 10:42:51

26

Views

Why does $M \equiv a^{p-1}M\pmod p$ implies $a^{p-1}\equiv 1\pmod p$?

Published on 03 Apr 2018 - 10:42

score 2.8k · Answer 4 · 2018-04-03 19:12:59

2.8k

Views

Find the number of solutions to a congruence equation

Published on 03 Apr 2018 - 19:12

score 263 · Answer 5 · 2018-04-04 20:31:05

263

Views

Proof that $\sum\limits_i \frac{(p-1)!}i$ is divisible by $p$

Published on 04 Apr 2018 - 20:31

score 324 · Answer 6 · 2018-04-06 23:10:46

324

Views

Number of solutions to a congruence equation

Published on 06 Apr 2018 - 23:10

score 103 · Answer 7 · 2026-03-24 07:21:43

103

Views

Prove that $a^2,a^4,...,a^{p-1}$ are the quadratic residues modulo $p$.

Published on 24 Mar 2026 - 7:21

score 580 · Answer 8 · 2018-04-08 02:10:51

580

Views

How many integers in the range satisfy a given congruence?

Published on 08 Apr 2018 - 2:10

score 65 · Answer 9 · 2018-04-09 21:43:24

65

Views

Show that there is a unique $k$ such that $1\le k\le5$ and $5\mid b+2^nk$ for any positive integers $b$ and $n$.

Published on 09 Apr 2018 - 21:43

score 988 · Answer 10 · 2018-04-10 22:24:59

988

Views

If $p$ is prime and $p$ $\equiv$ $1$ (mod 4), then the congruence $x^2$ $\equiv$ $-1$ (mod $p$) has two incongruent solutions...

Published on 10 Apr 2018 - 22:24

score 136 · Answer 11 · 2018-04-12 03:42:17

136

Views

Find all $x \in \mathbb{Z}$ satisfying the congruence equation $x \equiv 1 \pmod 5$

Published on 12 Apr 2018 - 3:42

score 65 · Answer 12 · 2026-03-25 06:57:38

65

Views

If n = $\frac{a^{2p}-1}{a^2-1}$ where $a$ is an integer $>1$ and $p$ is an odd prime, then $n$ is pseudoprime to the base $a$.

Published on 25 Mar 2026 - 6:57

score 238 · Answer 13 · 2018-04-13 03:50:25

238

Views

$x^2 \equiv -1$ (mod $p$) has two incongruent solutions given by $x\equiv\pm(\frac{p-1}{2})!$(mod $p$)

Published on 13 Apr 2018 - 3:50

score 28 · Answer 14 · 2018-04-13 12:34:08

28

Views

What does $(k!)^2((n-1)!+1)+n(k!-1)(k-1)!\equiv0$ mod $n(n+k)$ reduce/simplify into?

Published on 13 Apr 2018 - 12:34

score 201 · Answer 15 · 2026-03-25 04:41:07

201

Views

Prove that $24n+5$ can't be a perfect square with $n \in \Bbb Z^+$

Published on 25 Mar 2026 - 4:41

List Question

How to solve the below system of congruence?

Odd integer is power of 3 modulo power of 2

Why does $M \equiv a^{p-1}M\pmod p$ implies $a^{p-1}\equiv 1\pmod p$?

Find the number of solutions to a congruence equation

Proof that $\sum\limits_i \frac{(p-1)!}i$ is divisible by $p$

Number of solutions to a congruence equation

Prove that $a^2,a^4,...,a^{p-1}$ are the quadratic residues modulo $p$.

How many integers in the range satisfy a given congruence?

Show that there is a unique $k$ such that $1\le k\le5$ and $5\mid b+2^nk$ for any positive integers $b$ and $n$.

If $p$ is prime and $p$ $\equiv$ $1$ (mod 4), then the congruence $x^2$ $\equiv$ $-1$ (mod $p$) has two incongruent solutions...

Find all $x \in \mathbb{Z}$ satisfying the congruence equation $x \equiv 1 \pmod 5$

If n = $\frac{a^{2p}-1}{a^2-1}$ where $a$ is an integer $>1$ and $p$ is an odd prime, then $n$ is pseudoprime to the base $a$.

$x^2 \equiv -1$ (mod $p$) has two incongruent solutions given by $x\equiv\pm(\frac{p-1}{2})!$(mod $p$)

What does $(k!)^2((n-1)!+1)+n(k!-1)(k-1)!\equiv0$ mod $n(n+k)$ reduce/simplify into?

Prove that $24n+5$ can't be a perfect square with $n \in \Bbb Z^+$

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