Prove-that-there-is-no-solution-for-7-n-1-mod-35-with-n-N-

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