Find all solutions of the form (a,b) to the equation:
a! + 8 = 2b
such that a,b∊N (i.e., a and b are natural numbers).
(Australia)
Solution
First we note that a! ≡ 0(mod 16) for a ≥ 6 and 2b ≡ 0(mod 16) for b ≥ 4.
For a ≥ 6, the equation in mod 16 becomes
a! + 8 ≡ 0 + 8 ≡ 8 (mod 16).
Since a! > 24 for a ≥ 6, the left hand side and right hand side are different in mod 16 with
a! + 8 ≡ 0 + 8 ≡ 8 (mod 16), and 2b ≡ 0(mod 16).
From this observation, it is easy to see that the only solutions to this Diophantine equation must have a < 6. On checking for a = 1, 2, 3, 4, 5, we find that the only solutions (a,b) are (4,5) and (5,7).
Comments
Post a Comment