Prove that equation
2x-y! = 2
Has only 2 valid integer solutions.
Bonus: How about x2-y! = 2 ?
(In reply to
Puzzle Solution: Part A by K Sengupta)
Since 3!=6, it follows that:
y!= 0 ( mod 3) whenever y>=3
From the given relation, x^2-y! = 2, we must have:
Then,
x^2= 2 (mod 3) corresponding to y>=3
BUT, 2 is NOT a quadratic residue in the mod 3 system.
This is a contradiction.
Accordingly, y=1, 2
If y=1, then x^2=3, giving x =+/- V3, a contradiction.
If y=2, then, x^2=4 giving x= +/-2
Consequently,
(x, y) =(2, 2), (-2, 2) are the only possible pairs in consonance with the given conditions.
Edited on June 8, 2022, 7:44 am
Puzzle Solution: Part B
