Prove that equation
2x-y! = 2
Has only 2 valid integer solutions.
Bonus: How about x2-y! = 2 ?
(In reply to
Puzzle Solution: Part B by K Sengupta)
x^2-y! = 2
or, y!= x^2 -2
Since, 5!=120, it follows that:
x^2=2(mod 10) whenever, y>= 5
Since 2 is NOT a quadratic residue in the mod 10 system, this leads to a contradiction.
Therefore, n integer solution is possible whenever y>= 5.
Now, checking for y=1, 2, 3, and 4 - we observe that:
y=2 gives x^2 = 2!+2 =4, so that: x=2, -2
None of the other three values of y lead to an integer value of x.
Consequently, (x, y) =(2, 2), (-2, 2) are the only possible solutions.
Puzzle Solution: Part B : Method 2
