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From factorial to power (Posted on 2020-08-31) Difficulty: 3 of 5
There are two distinct integer solutions to


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

No Solution Yet Submitted by Ady TZIDON    
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Solution Solution Comment 1 of 1
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:

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.
  Posted by Larry on 2020-08-31 08:55:06
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