Prove that there is no solution for 7^n ≡1 mod(35) with n∈N.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 80643 by mr W last updated on 05/Feb/20

$P r o v e t h a t t h e r e i s n o s o l u t i o n f o r$ $7^{n} \equiv 1 m o d (35) w i t h n \in N .$
	Answered by MJS last updated on 05/Feb/20

	$\forall k \in N : 7^{20 k} \equiv 1 \mod 55$
	Commented by mr W last updated on 05/Feb/20

	$t h a n k s s i r! i s h o u l d h a v e a s k e d$ $7^{n} \equiv 1 m o d (35) . i s t h e r e a s o l u t i o n ?$
	Commented by MJS last updated on 05/Feb/20

	$no solution except n = 0$ $k \in N^{★} : 7^{4 k} \equiv 21 \mod 35$ $k \in N :$ $7^{4 k + 1} \equiv 7 \mod 35$ $7^{4 k + 2} \equiv 14 \mod 35$ $7^{4 k + 3} \equiv 28 \mod 35$
	Commented by mr W last updated on 05/Feb/20

	$t h a n k y o u s i r f o r c o n f i r m i n g!$
	Answered by mind is power last updated on 05/Feb/20

	$o n l y n = 0$ $i f n ⩾ 1 \Rightarrow 7 ∣ 7^{n}, 7 ∣ 35$ $i f \exists n s u c h 7^{n} = 1 (35) \Rightarrow$ $7^{n} = 1 + 35 k \Rightarrow$ $1 = 7 (7^{n - 1} - 5 k) \Rightarrow 7 ∣ 1 a b s u r d$
	Commented by mr W last updated on 05/Feb/20

	$t h a n k s a l o t s i r! {t h a t}^{'} s i t!$
	Commented by mind is power last updated on 05/Feb/20

	$y^{'} r e W e l c o m$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum