Question

Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$?

score 133 · Answer 1 · 2019-09-27 12:35:52

133

Views

Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$?

Published on 27 Sep 2019 - 12:35

score 125 · Answer 2 · 2019-09-30 18:21:21

125

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Help understand this gcd proof for $gcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b − a) = gcd(a, b + a)$

Published on 30 Sep 2019 - 18:21

score 276 · Answer 3 · 2014-03-16 15:52:56

276

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Proof for: GCD and divisibility

Published on 16 Mar 2014 - 15:52

score 8.2k · Answer 4 · 2014-03-18 07:47:49

8.2k

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Proof $ GCD(a,b) = GCD(a, b-a) = GCD (a, r_b) $

Published on 18 Mar 2014 - 7:47

score 4.3k · Answer 5 · 2014-03-19 20:24:10

Show $GCD(a_1, a_2, a_3, \ldots , a_n)$ is the least positive integer that can be expressed in the form $a_1x_1+a_2x_2+ \ldots +a_nx_n$

score 173 · Answer 6 · 2014-03-26 05:00:58

173

Views

The inverse of m with respect to n in modular arithmetics

Published on 26 Mar 2014 - 5:00

score 824 · Answer 7 · 2014-04-06 06:11:23

824

Views

Why if $a = qb + r$, then $\operatorname{gcd}(a,b) = \operatorname{gcd}(b, r)$ intuitively?

Published on 06 Apr 2014 - 6:11

score 215 · Answer 8 · 2014-04-08 15:40:48

215

Views

Prove that the integers $n$ and $n, 2^{2^n} + 1$ are relatively prime.

Published on 08 Apr 2014 - 15:40

score 2.5k · Answer 9 · 2014-04-14 05:43:37

2.5k

Views

Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$

Published on 14 Apr 2014 - 5:43

score 7.1k · Answer 10 · 2014-04-18 08:06:01

7.1k

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Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$

Published on 18 Apr 2014 - 8:06

score 731 · Answer 11 · 2014-04-18 20:27:02

731

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If $\gcd(a, b) = 1$ then prove that $\gcd(a+b, a^2-ab+b^2) = 1$ or $3$?

Published on 18 Apr 2014 - 20:27

score 579 · Answer 12 · 2014-04-22 02:57:09

579

Views

Prove:$a|bc$ and $\gcd(a,b)=1$ implies $a|c$

Published on 22 Apr 2014 - 2:57

score 156 · Answer 13 · 2014-04-25 20:46:34

156

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How to prove that $\gcd(2n^3-14n+2,n+3) = \gcd(n+3,10)$?

Published on 25 Apr 2014 - 20:46

score 358 · Answer 14 · 2014-05-04 03:06:18

358

Views

Let $x$ and $y$ be positive integers, $x < y$, and $x + y = 667$. Find all pairs $(x,y)$ if $\text{lcm}\,(x,y)/\gcd\,(x,y) = 120$.

Published on 04 May 2014 - 3:06

score 79 · Answer 15 · 2014-05-05 04:25:42

79

Views

Can we characterize the pairs of positive integers $b<a$ such that: $b|(a^2+\gcd(a,b))$

Published on 05 May 2014 - 4:25

List Question

Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$?

Help understand this gcd proof for $gcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b − a) = gcd(a, b + a)$

Proof for: GCD and divisibility

Proof $ GCD(a,b) = GCD(a, b-a) = GCD (a, r_b) $

Show $GCD(a_1, a_2, a_3, \ldots , a_n)$ is the least positive integer that can be expressed in the form $a_1x_1+a_2x_2+ \ldots +a_nx_n$

The inverse of m with respect to n in modular arithmetics

Why if $a = qb + r$, then $\operatorname{gcd}(a,b) = \operatorname{gcd}(b, r)$ intuitively?

Prove that the integers $n$ and $n, 2^{2^n} + 1$ are relatively prime.

Proving $\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1$

Prove for all $n \in N$, $\gcd(2n+1,9n+4)=1$

If $\gcd(a, b) = 1$ then prove that $\gcd(a+b, a^2-ab+b^2) = 1$ or $3$?

Prove:$a|bc$ and $\gcd(a,b)=1$ implies $a|c$

How to prove that $\gcd(2n^3-14n+2,n+3) = \gcd(n+3,10)$?

Let $x$ and $y$ be positive integers, $x < y$, and $x + y = 667$. Find all pairs $(x,y)$ if $\text{lcm}\,(x,y)/\gcd\,(x,y) = 120$.

Can we characterize the pairs of positive integers $b<a$ such that: $b|(a^2+\gcd(a,b))$

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