Prove that equation
2x-y! = 2
Has only 2 valid integer solutions.
Bonus: How about x2-y! = 2 ?
2^x -y! =2
or, y=2^x - 2
For y=1, we have:
2^x=3, which is a contradiction.
For y=2, we have:
2^x = 2! +2 = 4
=> x=2
For y=3, we have:
2^x =3!+2 =8
=> x=3
For y>= 4, we have:
y!= 0 (mod 8)
or, 2^x = 2 (mod 8)....(#l)
Now, we observe that
2^1=2( mod 8), 2^2= 4 (mod 8)
and, 2^x= 0 (mod 8) for x>=3
Therefore,, x should equal 1
But at x=1, we have:
y!=2^1-2 =0, which is a contradiction.
Accordingly, we must have:
(x, y) = (2, 2), (3, 3) are the only integer pairs in conformity with the given conditions.
This PROVES that there are precisely 2 VALID integer solutions.
Edited on June 8, 2022, 7:10 am
Puzzle Solution: Part A
