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 From factorial to power (Posted on 2020-08-31)
There are two distinct integer solutions to

x!+8=2y

a. Find both.
b. Prove no other solutions exist.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Solution | Comment 1 of 2
part (a)
(x,y) = {(4,5), (5,7)}

For this I used the following short Python program which found only the 2 results.

import math
for x in range(10000):
y = math.log( (math.factorial(x) + 8) , 2)
if int(round(y)) == y:
print(x,y)

part (b)
Proof:   Start with last solution of (x=5,y=7) and suppose there is a next larger solution (X, Y).
To be integers, X and Y must be
X = 5+a
Y = 7+b

(5+a)! + 8 = 2^7 * 2^b
divide both sides by 8
(5+a)!/8 + 1 = 2^4 * 2^b  which is true if (a,b) = (0,0)
for a>0, (5+a)!/8 always ends in zero, so (5+a)!/8 + 1 ends in one.
But 2^4 * 2^b never ends in one.
qed
 Posted by Larry on 2020-08-31 08:55:06

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