The claim is true. This is a somewhat lacklustre proof:
1. Check that the proposition seems true using small values:
Start with the idea that floor(4^n  (2+《qrt(2))^n) = (4^n  (2+《qrt(2))^n  (2sqrt(2))^n), the Pelltype expression being more convenient to manipulate.
2. Then a = (4^n  (2+《qrt(2))^n  (2sqrt(2))^n)/112, 0<n<20 produces: {{a == 0, n == 1}, {a == 5, n == 5}, {a == 98, n == 7}, {a == 30888, n == 11}, {a == 522704, n == 13}, {a == 142999104, n == 17}, {a == 2333122176, n == 19}}
3. Looking at 4^n, it is easy to see that it is periodic mod112 with period 3 for n>1 and values: {16,64,32}.
4. Looking at a = (2+《qrt(2))^n +(2sqrt(2))^n, 0<n<10 produces:
{{n == 1, x == 4}, {n == 2, x == 12}, {n == 3, x == 40}, {n == 4, x == 136}, {n == 5, x == 464}, {n == 6, x == 1584}, {n == 7, x == 5408}, {n == 8, x == 18464}, {n == 9, x == 63040}}.
5. By inspection, the integral recurrence relation for a is then clearly: a(n)= 4*a(n1)2*a(n2), and we can fill up an Excel sheet with 'small' values. (To save space, this exercise is left to the reader).
6. It transpires that after a bumpy start, the expression (2+《qrt(2))^n +(2sqrt(2))^n also settles down to period 6, mod112 with values: {16,16,32,96,96,80}. The two periods coincide to produce divisibility by 112 whenever n={6k1},{6k+1}, but not otherwise.
7. Since 112=28*4 I strongly suspect that a more 'profound' solution exists, possibly along the lines of 'Quarting the Cube' (pid8287); unfortunately I was unable to identify the clinching expressions.
Edited on October 25, 2013, 1:45 pm

Posted by broll
on 20131025 13:33:06 