Prove that equation
2x-y! = 2
Has only 2 valid integer solutions.
Bonus: How about x2-y! = 2 ?
First part:
2^x-y! = 2 or 2^x = y! + 2
All factorials above 3! are 0 mod 4.
For all integers x > 1, 2^x is also 0 mod 4.
But then when x > 1, (2^x- 2) is 2 mod 4.
(and negative x results in fractions)
So all factorials are eliminated except 0!, 1!, 2!, 3!
So the only (x,y) pairs that solve the equation are:
(2,2) 2^2 - 2! = 4 - 2 = 2, and
(3,3) 2^3 - 3! = 8 - 6 = 2 respectively.
Bonus
x^2-y! = 2
Squares are all either 0 or 1 mod 3.
Squares minus 2 are either 1 or 2 mod 3.
But Factorials above 2! are all 0 mod 3.
So again we are limited to only a few factorials: 0!, 1!, 2! = 1, 1, 2
So the only (x,y) pairs are:
(2,2) 2^2 - 2! = 4 - 2 = 2
(-2,2) (-2)^2 - 2! = 4 - 2 = 2
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Posted by Larry
on 2022-06-08 07:36:38 |
Solution
