find-the-integers-x-y-z-n-that-satisfy-2-n-x-y-z-

Question Number 138989 by metamorfose last updated on 20/Apr/21

f i n d t h e i n t e g e r s x, y, z, n t h a t

s a t i s f y : 2^{n} = x! + y! + z!

Answered by mindispower last updated on 21/Apr/21

l e t m = m i n (x, y, z), i f m ⩾ 3 \Rightarrow

x! + y! + z! \equiv 0 [3] \Leftrightarrow 2^{n} \equiv 0 [3] i m p o s s i b s i n c e

2, 3 a r e c o p r i m e

\Rightarrow m \in {0, 1, 2} b y s y m e t r i e o f x! + y! + z!

l e t m = x, x ⩽ y ⩽ z

x = 0, 1 \Rightarrow 2^{n} - 1 = z! + y! ⩾ 2 \Rightarrow n ⩾ 2

z! + y! \equiv 1 [2] \Rightarrow y = {0, 1}

\Rightarrow 2^{n} - 2 = z! \Rightarrow 2 (2^{n - 1} - 1) = z!, w e s e e z ⩾ 2

s i n c e 2 ∣ 2 (2^{n - 1} - 1)

t h e p o w e r o f 2 i n t h e 2 (2^{n - 1} - 1) i s o n e s i n c e

2^{n - 1} - 1 \equiv 1 [2] \Rightarrow z < 4 \Rightarrow z \in {2, 3}

z = 2 \Rightarrow 2^{n} - 2 = 2 \Rightarrow n = 2

z = 3 \Rightarrow 2^{n} = 8 \Rightarrow n = 3

c a s e 2 x = 2

\Rightarrow 2^{n} - 2 = y! + z! \Rightarrow 2 (2^{n - 1} - 1) = y! + z!, 2 ⩽ y < 4

b e c a u s e p o w e r o f 2 i n 2 (2^{n - 1} - 1) = 1

a n d y ⩽ z i f y ⩾ 4 \Rightarrow 4 ∣ (y! + z!) \Rightarrow p o w e r o f 2 i n

y! + z! i s a t l e s s 2 i m p o s s i b l

i f y = 2

\Rightarrow 2^{2} (2^{n - 2} - 1) = z! \Rightarrow n o s o l u t i o n

y = 3 \Rightarrow 2 (2^{n - 1} - 1) = 6 + z!

\Rightarrow 2^{3} (2^{n - 3} - 1) = z! \Rightarrow z = {4, 5}

z = 4 \Rightarrow 24 = 2^{n} - 8 \Rightarrow n = 5

z = 5 \Rightarrow 128 = 2^{n} \Rightarrow n = 7, a n d e r p e r m u t a t i o n

o f x, y, z

Commented by metamorfose last updated on 21/Apr/21

t h n x s i r