find the integers x , y , z , n that satisfy : 2^n =x!+y!+z!


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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
	$let m=min(x,y,z),if m≥3⇒ x!+y!+z!≡0[3]⇔2^n ≡0[3] impossib since 2,3 are coprime ⇒m∈{0,1,2} by symetrie of x!+y!+z! let m=x,x≤y≤z x=0,1⇒2^n −1=z!+y!≥2⇒n≥2 z!+y!≡1[2]⇒y={0,1} ⇒2^n −2=z!⇒2(2^(n−1) −1)=z!,we see z≥2 since 2∣2(2^(n−1) −1) the power of 2 in the 2(2^(n−1) −1)is one since 2^(n−1) −1≡1[2]⇒z<4⇒z∈{2,3} z=2⇒2^n −2=2⇒n=2 z=3⇒2^n =8⇒n=3 case 2 x=2 ⇒2^n −2=y!+z!⇒2(2^(n−1) −1)=y!+z!,2≤y<4 because power of 2 in 2(2^(n−1) −1)=1 and y≤z if y≥4⇒4∣(y!+z!) ⇒power of 2 in y!+z! is at less 2 impossibl if y=2 ⇒2^2 (2^(n−2) −1)=z!⇒no solution y=3⇒2(2^(n−1) −1)=6+z! ⇒2^3 (2^(n−3) −1)=z!⇒z={4,5} z=4⇒24=2^n −8⇒n=5 z=5⇒128=2^n ⇒n=7,ander permutation of x,y,z$
	$l e t m = m i n (x, y, z), i f m ⩾ 3 \Rightarrow$ $x! + y! + z! \equiv 0 [3] \Leftrightarrow 2^{n} \equiv 0 [3] i m p o s s i b s i n c e$ $2, 3 a r e c o p r i m e$ $\Rightarrow m \in {0, 1, 2} b y s y m e t r i e o f x! + y! + z!$ $l e t m = x, x ⩽ y ⩽ z$ $x = 0, 1 \Rightarrow 2^{n} - 1 = z! + y! ⩾ 2 \Rightarrow n ⩾ 2$ $z! + y! \equiv 1 [2] \Rightarrow y = {0, 1}$ $\Rightarrow 2^{n} - 2 = z! \Rightarrow 2 (2^{n - 1} - 1) = z!, w e s e e z ⩾ 2$ $s i n c e 2 ∣ 2 (2^{n - 1} - 1)$ $t h e p o w e r o f 2 i n t h e 2 (2^{n - 1} - 1) i s o n e s i n c e$ $2^{n - 1} - 1 \equiv 1 [2] \Rightarrow z < 4 \Rightarrow z \in {2, 3}$ $z = 2 \Rightarrow 2^{n} - 2 = 2 \Rightarrow n = 2$ $z = 3 \Rightarrow 2^{n} = 8 \Rightarrow n = 3$ $c a s e 2 x = 2$ $\Rightarrow 2^{n} - 2 = y! + z! \Rightarrow 2 (2^{n - 1} - 1) = y! + z!, 2 ⩽ y < 4$ $b e c a u s e p o w e r o f 2 i n 2 (2^{n - 1} - 1) = 1$ $a n d y ⩽ z i f y ⩾ 4 \Rightarrow 4 ∣ (y! + z!) \Rightarrow p o w e r o f 2 i n$ $y! + z! i s a t l e s s 2 i m p o s s i b l$ $i f y = 2$ $\Rightarrow 2^{2} (2^{n - 2} - 1) = z! \Rightarrow n o s o l u t i o n$ $y = 3 \Rightarrow 2 (2^{n - 1} - 1) = 6 + z!$ $\Rightarrow 2^{3} (2^{n - 3} - 1) = z! \Rightarrow z = {4, 5}$ $z = 4 \Rightarrow 24 = 2^{n} - 8 \Rightarrow n = 5$ $z = 5 \Rightarrow 128 = 2^{n} \Rightarrow n = 7, a n d e r p e r m u t a t i o n$ $o f x, y, z$
	Commented by metamorfose last updated on 21/Apr/21

	$t h n x s i r$
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