List questions from congruences

Published on 27 Nov 2013 - 15:58

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Linear congruence proof, show congruence has exactly two incongruent solutions

Published on 29 Nov 2013 - 1:28

87

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Solving linear congruences

Published on 30 Nov 2013 - 16:21

88

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How do you prove set with modulo?

Published on 30 Nov 2013 - 17:52

61

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Showing an induction step for a congruence relation.

Published on 02 Dec 2013 - 21:23

73

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Congruence equation problem [verification]

Published on 05 Dec 2013 - 17:57

65

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Is $20+k^2\not\equiv4(\text{mod}~7)$ for all $k\geq0$?

Published on 05 Dec 2013 - 20:22

65

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Why doesn't this proof that $a \equiv b \pmod{m}$ for any integers $a, b, m$ work?

Published on 06 Dec 2013 - 2:04

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If $(m,n)=1$ then $m^{\phi(n)}+n^{\phi(m)}\equiv0\pmod {mn}$

Published on 27 Mar 2026 - 23:48

388

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I suppose $(a,m)=(a-1,m)=1$, show that $1+a+a^2+\ldots+a^{\phi(m)-1}\equiv0\pmod m$

Published on 30 Mar 2026 - 7:09

348

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Congruence equation proof

Published on 06 Dec 2013 - 19:39

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Let $a,m\in\mathbb{N^*}$, with $m>1$, such that $(a,m)=1$. Show that, if $n_1\equiv n_2\pmod{\phi(m)}$, then $a^{n_1}\equiv a^{n_2}\pmod m$

Published on 30 Mar 2026 - 7:09

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Exponents and mod arithmetic

Published on 08 Dec 2013 - 3:03

2.2k

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system of linear congruences when moduli are not coprime

Published on 08 Dec 2013 - 9:26

868

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Finding mod of X^2+1 = 0 to have exactly 4 solutions

Published on 10 Dec 2013 - 14:29

List Question

${{a+b\sqrt-7}\over{2}}$ as the root of a polynomial

Linear congruence proof, show congruence has exactly two incongruent solutions

Solving linear congruences

How do you prove set with modulo?

Showing an induction step for a congruence relation.

Congruence equation problem [verification]

Is $20+k^2\not\equiv4(\text{mod}~7)$ for all $k\geq0$?

Why doesn't this proof that $a \equiv b \pmod{m}$ for any integers $a, b, m$ work?

If $(m,n)=1$ then $m^{\phi(n)}+n^{\phi(m)}\equiv0\pmod {mn}$

I suppose $(a,m)=(a-1,m)=1$, show that $1+a+a^2+\ldots+a^{\phi(m)-1}\equiv0\pmod m$

Congruence equation proof

Let $a,m\in\mathbb{N^*}$, with $m>1$, such that $(a,m)=1$. Show that, if $n_1\equiv n_2\pmod{\phi(m)}$, then $a^{n_1}\equiv a^{n_2}\pmod m$

Exponents and mod arithmetic

system of linear congruences when moduli are not coprime

Finding mod of X^2+1 = 0 to have exactly 4 solutions

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