There are two distinct integer solutions to
x!+8=2^{y}
a. Find both.
b. Prove no other solutions exist.
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 20200831 08:55:06 