Question

Number Theory: Divisibility on a set of consecutive integers

score 80 · Answer 1 · 2026-04-15 19:35:17

80

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Number Theory: Divisibility on a set of consecutive integers

Published on 15 Apr 2026 - 19:35

score 65 · Answer 2 · 2026-04-16 00:43:28

65

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Prove that there are no two integers $a$ and $b$, such that $\frac{a}{b}$ is non-terminating and non-recurring number

Published on 16 Apr 2026 - 0:43

score 163 · Answer 3 · 2026-04-15 06:09:28

163

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Number theory: Divisibility in block of any size

Published on 15 Apr 2026 - 6:09

score 77 · Answer 4 · 2026-04-15 19:46:19

77

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A $p$-congruence involving both kinds of Stirling numbers.

Published on 15 Apr 2026 - 19:46

score 156 · Answer 5 · 2026-04-23 15:47:21

156

Views

Prove that,if $n$ is an integer, then $n^2 + 15n = 1$ is odd

Published on 23 Apr 2026 - 15:47

score 413 · Answer 6 · 2026-04-15 23:54:41

413

Views

$x^k\equiv1 \pmod{p}$ for all nonzero $x$, then $p-1$ divides $k$

Published on 15 Apr 2026 - 23:54

score 154 · Answer 7 · 2026-04-16 23:21:53

154

Views

prove that there exists no integer in $\pm\frac1k \pm \frac1{k+1}\pm\frac1{k+2}...\pm\frac1{k+n} $

Published on 16 Apr 2026 - 23:21

score 48 · Answer 8 · 2026-04-10 22:03:59

48

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Multiplying factors of a positive number

Published on 10 Apr 2026 - 22:03

score 161 · Answer 9 · 2026-04-17 14:01:58

161

Views

Given a prime $p = 2^n+1$, show that $\alpha^{(p-1)/2} \equiv -1 \pmod p$

Published on 17 Apr 2026 - 14:01

score 387 · Answer 10 · 2026-04-10 17:17:27

387

Views

$n^2 (n^4 - 1)$ is divisible by 60

Published on 10 Apr 2026 - 17:17

score 74 · Answer 11 · 2026-04-16 14:52:19

74

Views

Why does $6^{2m}$ mod 7 $\not\equiv 6^{2n + 1}$ mod 7 for all integers $m$ and $n$?

Published on 16 Apr 2026 - 14:52

score 110 · Answer 12 · 2026-03-25 09:32:24

110

Views

Completing the Square with a quartic polynomial

Published on 25 Mar 2026 - 9:32

score 125 · Answer 13 · 2026-04-16 04:30:24

125

Views

If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?

Published on 16 Apr 2026 - 4:30

score 154 · Answer 14 · 2026-04-16 08:18:28

154

Views

p is prime. Show that there is at most one $(m,n) \in \mathbb Z^2$ with $0<m\le n$ that solves $p = m^2 + n^2$

Published on 16 Apr 2026 - 8:18

score 171 · Answer 15 · 2026-04-19 22:35:14

171

Views

$2^{3^{4^{...^{n}}}} \equiv 1$ (mod $n+1$)

Published on 19 Apr 2026 - 22:35

List Question

Number Theory: Divisibility on a set of consecutive integers

Prove that there are no two integers $a$ and $b$, such that $\frac{a}{b}$ is non-terminating and non-recurring number

Number theory: Divisibility in block of any size

A $p$-congruence involving both kinds of Stirling numbers.

Prove that,if $n$ is an integer, then $n^2 + 15n = 1$ is odd

$x^k\equiv1 \pmod{p}$ for all nonzero $x$, then $p-1$ divides $k$

prove that there exists no integer in $\pm\frac1k \pm \frac1{k+1}\pm\frac1{k+2}...\pm\frac1{k+n} $

Multiplying factors of a positive number

Given a prime $p = 2^n+1$, show that $\alpha^{(p-1)/2} \equiv -1 \pmod p$

$n^2 (n^4 - 1)$ is divisible by 60

Why does $6^{2m}$ mod 7 $\not\equiv 6^{2n + 1}$ mod 7 for all integers $m$ and $n$?

Completing the Square with a quartic polynomial

If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?

p is prime. Show that there is at most one $(m,n) \in \mathbb Z^2$ with $0<m\le n$ that solves $p = m^2 + n^2$

$2^{3^{4^{...^{n}}}} \equiv 1$ (mod $n+1$)

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