Question

How to solve the below system of congruence?

score 52 · Answer 1 · 2026-04-10 03:57:48

52

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How to solve the below system of congruence?

Published on 10 Apr 2026 - 3:57

score 51 · Answer 2 · 2026-04-14 10:30:50

51

Views

Odd integer is power of 3 modulo power of 2

Published on 14 Apr 2026 - 10:30

score 29 · Answer 3 · 2026-04-17 02:13:43

29

Views

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

Published on 17 Apr 2026 - 2:13

score 2.8k · Answer 4 · 2026-04-11 23:17:57

2.8k

Views

Find the number of solutions to a congruence equation

Published on 11 Apr 2026 - 23:17

score 266 · Answer 5 · 2026-04-11 06:04:59

266

Views

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

Published on 11 Apr 2026 - 6:04

score 328 · Answer 6 · 2026-05-10 15:02:29

328

Views

Number of solutions to a congruence equation

Published on 10 May 2026 - 15:02

score 106 · Answer 7 · 2026-05-06 23:55:19

106

Views

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

Published on 06 May 2026 - 23:55

score 583 · Answer 8 · 2026-04-12 14:53:35

583

Views

How many integers in the range satisfy a given congruence?

Published on 12 Apr 2026 - 14:53

score 68 · Answer 9 · 2026-04-16 06:45:37

68

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 16 Apr 2026 - 6:45

score 991 · Answer 10 · 2026-04-16 02:32:16

991

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 16 Apr 2026 - 2:32

score 139 · Answer 11 · 2026-04-14 04:37:34

139

Views

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

Published on 14 Apr 2026 - 4:37

score 68 · Answer 12 · 2026-04-22 21:24:47

68

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 22 Apr 2026 - 21:24

score 241 · Answer 13 · 2026-04-15 20:47:30

241

Views

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

Published on 15 Apr 2026 - 20:47

score 31 · Answer 14 · 2026-04-14 17:50:15

31

Views

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

Published on 14 Apr 2026 - 17:50

score 204 · Answer 15 · 2026-05-10 15:09:29

204

Views

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

Published on 10 May 2026 - 15:09

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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