Question

How to show that $\gcd(n! + 1, (n + 1)! + 1) \mid n$?

score 3k · Answer 1 · 2010-10-09 17:00:17

3k

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How to show that $\gcd(n! + 1, (n + 1)! + 1) \mid n$?

Published on 09 Oct 2010 - 17:00

score 1.4k · Answer 2 · 2010-10-13 03:08:47

1.4k

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$b \mid ac\Rightarrow b \mid (a,b)(c,b)\,$ for integers $\,a,b,c$

Published on 13 Oct 2010 - 3:08

score 52.1k · Answer 3 · 2010-10-22 00:47:26

52.1k

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Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

Published on 22 Oct 2010 - 0:47

score 1.8k · Answer 4 · 2010-11-04 19:23:44

1.8k

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Computing GCD of all permutations (of the digits) of a given number

Published on 04 Nov 2010 - 19:23

score 11.9k · Answer 5 · 2010-11-15 14:21:37

11.9k

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$I+J=1 \Rightarrow I^n+J^n = 1$ for ideals (or elements in GCD domain) [Freshman's Dream Binomial Theorem]

Published on 15 Nov 2010 - 14:21

score 228 · Answer 6 · 2010-11-16 00:46:01

228

Views

$\gcd(bbb\dots b, bbb\dots b)$

Published on 16 Nov 2010 - 0:46

score 21.4k · Answer 7 · 2010-11-16 21:03:07

21.4k

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Order of elements is lcm-closed in abelian groups

Published on 16 Nov 2010 - 21:03

score 9.7k · Answer 8 · 2026-03-26 16:06:30

9.7k

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$\gcd(b^x - 1, b^y - 1, b^ z- 1,\dots) = b^{\gcd(x, y, z,\dots)} -1$

Published on 26 Mar 2026 - 16:06

score 3.6k · Answer 9 · 2010-12-13 00:16:06

3.6k

Views

When does the distributive law apply to ideals in a commutative ring?

Published on 13 Dec 2010 - 0:16

score 3.5k · Answer 10 · 2011-01-01 17:06:51

3.5k

Views

If $\gcd(a,b)=d$, then $\gcd(ac,bc)=cd$?

Published on 01 Jan 2011 - 17:06

score 613 · Answer 11 · 2011-02-01 04:16:08

613

Views

What's special about the greatest common divisor of a + b and a - b?

Published on 01 Feb 2011 - 4:16

score 2.3k · Answer 12 · 2011-02-02 17:28:08

2.3k

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$a|bc\!\!\iff\!\! a|(a,b)c\!\iff\!\!\frac{a}{(a,b)}\!\!\mid\! c\!\iff\!\!\frac{{\rm lcm}(a,b)}{b}\!\mid\! c\ $ [general Euclid's Lemma]

Published on 02 Feb 2011 - 17:28

score 39.7k · Answer 13 · 2011-02-07 22:27:51

39.7k

Views

Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $

Published on 07 Feb 2011 - 22:27

score 3.6k · Answer 14 · 2011-02-11 15:38:22

3.6k

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$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$

Published on 11 Feb 2011 - 15:38

score 13.8k · Answer 15 · 2011-02-23 03:06:02

13.8k

Views

Prove: $\gcd(a,b) = \gcd(a, b + at)$.

Published on 23 Feb 2011 - 3:06

List Question

How to show that $\gcd(n! + 1, (n + 1)! + 1) \mid n$?

$b \mid ac\Rightarrow b \mid (a,b)(c,b)\,$ for integers $\,a,b,c$

Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

Computing GCD of all permutations (of the digits) of a given number

$I+J=1 \Rightarrow I^n+J^n = 1$ for ideals (or elements in GCD domain) [Freshman's Dream Binomial Theorem]

$\gcd(bbb\dots b, bbb\dots b)$

Order of elements is lcm-closed in abelian groups

$\gcd(b^x - 1, b^y - 1, b^ z- 1,\dots) = b^{\gcd(x, y, z,\dots)} -1$

When does the distributive law apply to ideals in a commutative ring?

If $\gcd(a,b)=d$, then $\gcd(ac,bc)=cd$?

What's special about the greatest common divisor of a + b and a - b?

$a|bc\!\!\iff\!\! a|(a,b)c\!\iff\!\!\frac{a}{(a,b)}\!\!\mid\! c\!\iff\!\!\frac{{\rm lcm}(a,b)}{b}\!\mid\! c\ $ [general Euclid's Lemma]

Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $

$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$

Prove: $\gcd(a,b) = \gcd(a, b + at)$.

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