2^(m−1) =1+mn m, n ∈Z


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Question Number 211943 by Frix last updated on 25/Sep/24

$2^{m - 1} = 1 + m n$ $m, n \in Z$
Commented by BHOOPENDRA last updated on 25/Sep/24
${(m,n)∈Z×Z ∣ m is odd &n=((2^(m−1) −1)/m)} Example of solutions .m=1,n=0 .m=3,n=1 .m=7,n=9 The integer solution (m,n) for both equation exist only when m is odd integer that is not a multiple of 3 .specially ,for value of m that are multiples of P(prime)(Except 1) (i.e,m=P(2k+1) where k∈Z^+ ), P ≥3there are no integer solutions for n.$
${(m, n) \in Z \times Z ∣ m i s o d d & n = \frac{2^{m - 1} - 1}{m}}$ $E x a m p l e o f s o l u t i o n s$ $. m = 1, n = 0$ $. m = 3, n = 1$ $. m = 7, n = 9$ $T h e i n t e g e r s o l u t i o n (m, n) f o r$ $b o t h e q u a t i o n e x i s t o n l y w h e n m i s$ $o d d i n t e g e r t h a t i s n o t a m u l t i p l e o f$ $3 . s p e c i a l l y, f o r v a l u e o f m t h a t a r e$ $m u l t i p l e s o f P (p r i m e) (E x c e p t 1)$ $(i . e, m = P (2 k + 1) w h e r e k \in Z^{+}),$ $P \geq 3 t h e r e a r e n o i n t e g e r s o l u t i o n s f o r n .$
	Answered by BHOOPENDRA last updated on 25/Sep/24
	${(m,n)∈Z×Z ∣m is odd and n=((2^(m−1) −1)/m)} for all odd value of m ∈Z^+ But for values of m that are odd multiple of P(prime) (i.e m=P(2k+1) where k∈ Z^+ & k≥1 ) there is no integer solutions for n except 1. N={m∈Z^+ :m=P(2k+1),k∈Z,k≥1,P≥3} The n=((2^(m−1) −1)/(m )) for odd m. 1. 2^(m−1) =1+mn 2. 2^(m−1) =1−mn Previous results Recape (for Equation 1)S_1 .m=1 ,n=0 m=3 n=1 m=5 ,n=3 m=7 ,n=9 m=9 ,no solution(n=28.3333) m=15 no solution m=11 n=93 m=13 n=315 m=15 no solution m=17 n=3855 m=21 no solution m=25 no solution m=27 no solution m=33 no solution similarly for equation 2 S_2 ={(1,0),(3,−1),(5,−3),(11,−93).....} Key Insight m=p(2k+1) for P≥3 ,k≥1 there are no solutions for n (Except 1)$
	${(m, n) \in Z \times Z ∣ m i s o d d a n d n = \frac{2^{m - 1} - 1}{m}}$ $f o r a l l o d d v a l u e o f m \in Z^{+}$ $B u t f o r v a l u e s o f m t h a t a r e o d d$ $m u l t i p l e o f P (p r i m e)$ $(i . e m = P (2 k + 1) w h e r e$ $k \in Z^{+} & k ⩾ 1) t h e r e i s n o i n t e g e r s o l u t i o n s$ $f o r n e x c e p t 1 .$ $N = {m \in Z^{+} : m = P (2 k + 1), k \in Z, k ⩾ 1, P ⩾ 3}$ $T h e n = \frac{2^{m - 1} - 1}{m} f o r o d d m .$ $1 . 2^{m - 1} = 1 + m n$ $2 . 2^{m - 1} = 1 - m n$ $P r e v i o u s r e s u l t s R e c a p e$ $(f o r E q u a t i o n 1) S_{1}$ $. m = 1, n = 0$ $m = 3 n = 1$ $m = 5, n = 3$ $m = 7, n = 9$ $m = 9, n o s o l u t i o n (n = 28.3333)$ $m = 15 n o s o l u t i o n$ $m = 11 n = 93$ $m = 13 n = 315$ $m = 15 n o s o l u t i o n$ $m = 17 n = 3855$ $m = 21 n o s o l u t i o n$ $m = 25 n o s o l u t i o n$ $m = 27 n o s o l u t i o n$ $m = 33 n o s o l u t i o n$ $s i m i l a r l y f o r e q u a t i o n 2$ $S_{2} = {(1, 0), (3, - 1), (5, - 3), (11, - 93) . . . . .}$ $K e y I n s i g h t$ $m = p (2 k + 1) f o r P ⩾ 3, k ⩾ 1 t h e r e a r e$ $n o s o l u t i o n s f o r n (E x c e p t 1)$
	Commented by Frix last updated on 25/Sep/24

	$m = 9 \Rightarrow n = \frac{85}{3}$ $Check it again!$
	Commented by Frix last updated on 25/Sep/24

	$I first thought I made a mistake because$ $you got all odd m . . . I^{'} ve been in a hurry,$ $sorry for that .$
	Commented by BHOOPENDRA last updated on 25/Sep/24

	$C a n y o u c h e c k a g a i n i f t h e r e i s a n y$ $m i s t a k e i l l t h i n k a g a i n$
	Commented by Frix last updated on 25/Sep/24

	$So far {it}^{'} s ok . But also no solution for$ $m = 25, 35, 49, 55, . . .$ $Can we prove that m must be a prime$ $number ⩾ 3 ? (With the only exception of$ $m = 1)$
	Commented by A5T last updated on 25/Sep/24

	${F e r m a t}^{'} s t h e o r e m \Rightarrow 2^{p - 1} \equiv 1 (m o d p) (p \neq 2)$ $\Rightarrow p ∣ 2^{p - 1} - 1 \Rightarrow \frac{2^{p - 1} - 1}{p} \in Z .$ $S o, w h e n m i s p r i m e, n = \frac{2^{m - 1} - 1}{m} \in Z$ $B u t t h i s i s n o t s u f f i c i e n t t o c l a i m m (> 1) m u s t$ $a l w a y s b e a p r i m e . S o, i t s t i l l r e m a i n s t o$ $c o n s i d e r w h e r e m m a y b e c o m p o s i t e .$
	Commented by BHOOPENDRA last updated on 25/Sep/24

	$m = P (2 k + 1) w h e r e k \in Z^{+} a n d k ⩾ 1$ $P i s p r i m e P \geq 3 (e x c e p t 1)$ $w e c a n s a y t h a t i t h i n k$
	Answered by A5T last updated on 25/Sep/24

	$m, n \in Z \Rightarrow 1 + m n \in Z \Rightarrow m ⩾ 1 .$ $O t h e r w i s e, m (< 1) \Rightarrow 2^{m - 1} \notin Z$ $m = 1 \Rightarrow 1 = 1 + n \Rightarrow n = 0$ $L e t m > 1 a n d e v e n; t h e n 1 + m n i s o d d \to\leftarrow$ $S o, m > 1 a n d o d d, n = \frac{2^{m - 1} - 1}{m} \in Z \Rightarrow m ∣ 2^{m - 1} - 1$ $N o t e t h a t w h e n m i s p r i m e, 2^{m - 1} - 1 \equiv 0 (m o d m)$ $\Rightarrow F o r a l l p r i m e m, t h e r e e x i s t s n = \frac{2^{m - 1} - 1}{m} \in Z$ $F o r c o m p o s i t e :$ $s i n c e (2, m) = 1 \Rightarrow 2^{ϕ (m)} - 1 \equiv 0 (m o d m)$ $\Rightarrow ϕ (m) ∣ m - 1$ $S u p p o s e m i s c o m p o s i t e; a n d l e t t h e$ $f a c t o r i z a t i o n b e p_{1}^{e_{1}} p_{2}^{e_{2}} . . . p_{n}^{e_{n}} (p_{i} = p_{j} \Rightarrow i = j)$ $T h e n ϕ (m) = \underset{i = 1}{\prod^{n}} (p_{i}^{e_{i}} - p_{i}^{e_{i} - 1}) = \underset{i = 1}{\prod^{n}} p_{i}^{e_{i} - 1} (p_{i} - 1) ∣ m - 1$ $B u t s i n c e (m, m - 1) = 1 \Rightarrow e_{i} ⩽ 1$ $m = \underset{i = 1}{\prod^{n}} p_{i} \Rightarrow ϕ (m) = \underset{i = 1}{\prod^{n}} (p_{i} - 1) ∣ m - 1$ $L e t p_{n} b e t h e l a r g e s t p r i m e t h a t d i v i d e s m$ $\Rightarrow \frac{m}{p_{n}} = k \Rightarrow p_{n} k = m > m - 1$ $. . .$
	Commented by A5T last updated on 25/Sep/24

	$2^{ϕ (m)} \equiv 1 (m o d m) s i n c e g c d (2, m) = 1$ $B u t w e k n o w t h a t m ∣ 2^{m - 1} - 1, t h a t i s,$ $2^{m - 1} \equiv 1 (m o d m)$ $S i n c e ϕ (m) ⩽ m - 1; i t f o l l o w s t h a t ϕ (m) m u s t$ $d i v i d e m - 1 f r o m {L a g r a n g e}^{'} s t h e o r e m .$
	Commented by BHOOPENDRA last updated on 25/Sep/24

	$T h e k e y p o i n t i s t o r e c h e c k t h e a s s u m p t i o n$ $t h a t ϕ (m) ∣ m - 1 f o r c o m p o s i t e m .$ $T h i s i s n o t t r u e i n g e n e r a l, a n d t h u s$ $, t h e r e a s o n i n g f o r t h e c o m p o s i t e c a s e$ $b r e a k d o w n . w h i l e {f e r m a t}^{'} s L i t t l e t h e o r e m$ $f o r p r i m e, {E u l e r}^{'} s t h e o r e m d o e s n o t$ $g a u r a n t e e t h a t m ∣ 2^{m - 1} f o r c o m p o s i t e$ $m, b e a c u s e ϕ (m) d o e s n o t n e c e s s a r i l y$ $d i v i d e (m - 1)$ $T o c o r r e c t t h e r e a s o n i n g, w e w o u l d n e e d$ $t o c a r e f u l l y a n l y z e s p e c i f i c c o m p o s i t e$ $n u m b e r w h e r e m ∣ 2^{m - 1} h o l d s, w i t h o u t$ $r e l y i n g o n ϕ (m) ∣ m - 1 a s a g e n r a l r u l e .$
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