Question

If $p$ is a prime number and $a \in \mathbb{Z}$ isn't multiple of $p$, can we then state that the gcd(a,p)=1?

score 54 · Answer 1 · 2017-09-19 09:28:32

54

Views

If $p$ is a prime number and $a \in \mathbb{Z}$ isn't multiple of $p$, can we then state that the gcd(a,p)=1?

Published on 19 Sep 2017 - 9:28

score 108 · Answer 2 · 2017-09-19 14:50:36

108

Views

Does the equation $ax+by=\gcd(a,b)=\gcd(ax+by,b)=\gcd(a,ax+by)$ hold for $a, b, x, y \in \mathbb{Z}$?

Published on 19 Sep 2017 - 14:50

score 160 · Answer 3 · 2017-09-19 15:35:27

160

Views

how to prove: every two numbers gcd can be divided by this two numbers any common divisor

Published on 19 Sep 2017 - 15:35

score 162 · Answer 4 · 2017-09-20 17:21:15

162

Views

For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$

Published on 20 Sep 2017 - 17:21

score 147 · Answer 5 · 2017-09-21 13:47:32

147

Views

How to show $((a,b),b)=(a,b)$?

Published on 21 Sep 2017 - 13:47

score 2.3k · Answer 6 · 2017-09-23 23:51:21

2.3k

Views

Which of the following statements are true for all such $a$ and $b$? Prove the statement or give a counterexample.

Published on 23 Sep 2017 - 23:51

score 64 · Answer 7 · 2017-09-24 05:10:07

64

Views

Use $(a,b)=1$ and $a | bc$ then $a | c$ to show that if $p$ is prime and $a$ is an integer with $p | a^2$, then $p | a$

Published on 24 Sep 2017 - 5:10

score 233 · Answer 8 · 2017-09-26 04:19:36

233

Views

True or false : $\gcd(a_m, a_n) = a_{\gcd(m, n)}$?

Published on 26 Sep 2017 - 4:19

score 266 · Answer 9 · 2017-09-27 09:42:07

266

Views

$\gcd(p_{n-1}, \ n^5 - n^3 + n^2 - n + 1) = 1$ where $p_n = n$th prime.

Published on 27 Sep 2017 - 9:42

score 8.6k · Answer 10 · 2017-09-28 09:19:54

8.6k

Views

Prove $\gcd(ka,kb)=k\cdot\gcd(a,b)$

Published on 28 Sep 2017 - 9:19

score 1k · Answer 11 · 2017-09-29 18:38:57

1k

Views

Show that no GCD exists for $4$ and $2 \cdot (1 + \sqrt{-3})$ in $\{ x + y \sqrt{-3} | x, y \in \mathbb{Z}\}$

Published on 29 Sep 2017 - 18:38

score 1.8k · Answer 12 · 2017-10-02 02:59:27

1.8k

Views

Determining the number of distinct subgroups

Published on 02 Oct 2017 - 2:59

score 34 · Answer 13 · 2017-10-04 14:11:17

34

Views

Proof that sequence where two entries are the same, then the gcd of these indices must be the step

Published on 04 Oct 2017 - 14:11

score 62 · Answer 14 · 2017-10-04 14:26:02

62

Views

Greatest common divisor of two highly composite numbers

Published on 04 Oct 2017 - 14:26

score 2.5k · Answer 15 · 2017-10-05 00:28:09

2.5k

Views

If $a,b, c \in \mathbb{Z}$, then $c ·\gcd(a,b) ≤ gcd(ca, cb)$.

Published on 05 Oct 2017 - 0:28

List Question

If $p$ is a prime number and $a \in \mathbb{Z}$ isn't multiple of $p$, can we then state that the gcd(a,p)=1?

Does the equation $ax+by=\gcd(a,b)=\gcd(ax+by,b)=\gcd(a,ax+by)$ hold for $a, b, x, y \in \mathbb{Z}$?

how to prove: every two numbers gcd can be divided by this two numbers any common divisor

For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$

How to show $((a,b),b)=(a,b)$?

Which of the following statements are true for all such $a$ and $b$? Prove the statement or give a counterexample.

Use $(a,b)=1$ and $a | bc$ then $a | c$ to show that if $p$ is prime and $a$ is an integer with $p | a^2$, then $p | a$

True or false : $\gcd(a_m, a_n) = a_{\gcd(m, n)}$?

$\gcd(p_{n-1}, \ n^5 - n^3 + n^2 - n + 1) = 1$ where $p_n = n$th prime.

Prove $\gcd(ka,kb)=k\cdot\gcd(a,b)$

Show that no GCD exists for $4$ and $2 \cdot (1 + \sqrt{-3})$ in $\{ x + y \sqrt{-3} | x, y \in \mathbb{Z}\}$

Determining the number of distinct subgroups

Proof that sequence where two entries are the same, then the gcd of these indices must be the step

Greatest common divisor of two highly composite numbers

If $a,b, c \in \mathbb{Z}$, then $c ·\gcd(a,b) ≤ gcd(ca, cb)$.

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