The sequence an is defined by a1 = a2 = 1, an+2 = an+1 + 2an. The sequence bn is defined by b1 = 1, b2 = 7, bn+2 = 2bn+1 + 3bn. Show that the only integer in both sequences is 1.
Solution
We can solve the first recurrence relation to give an = A (-1)n + B 2n. Using a1 and a2, we get an = (2n + (-1)n+1)/3. Similarly, for the second recurrence relation we get bn = 2.3n-1 + (-1)n. So if am = bn then 2.3n + 3 (-1)n = 2m + (-1)m+1 or 2m = 2.3n + 3 (-1)n + (-1)m.
If m = 1 or 2, then we find n = 1 is the only solution, corresponding to the fact that the term 1 is in both sequences. If m > 2, then 2m = 0 mod 8. But 3n = (-1)n mod 4, so 2.3n + 3 (-1)n + (-1)m = 5 (-1)n + (-1)m mod 8 which cannot be 0 mod 8. So there are no solutions for m > 2.
© John Scholes
jscholes@kalva.demon.co.uk
19 Aug 2002