An approach using slightly lateral thinking:
N=4b^2-4
N=2a^2-2
(a^2-1)=2(b^2-1)
a^2-2b^2=1, for all qualifying N
Consider b: {1,5,29,169,985,..} A001653 in Sloane, in the context of the Pell Numbers, A000129 in Sloane:{1, 2, 5, 12, 29, 70, 169,..}
1^2+2^2 = 5
2^2+5^2 = 29
5^2+12^2 = 169
12^2+29^2 = 985
29^2+70^2 = 5741
70^2+169^2 = 33461
169^2+408^2 = 195025
...etc.
So each b is itself a sum of two squares, say x^2 and y^2.
We know that 96=2^5*3. We have 4(x^2+y^2)^2-4, certainly divisible by 4 ,for 2^2. But 4(x^2+y^2)^2-4 = 4 (x^2+y^2-1) (x^2+y^2+1); since successive Pell numbers have alternate parity, both brackets must be even, for another 2^2, and one must be divisible by 4, for the final power of 2.
Because x^2 and y^2 are squares, each is worth 0 or 1, mod 3. The Pell squares are periodic, mod 3: {1,1,1,0}. So both squares are worth 1, mod 3, or exactly one is worth 0.
Assume both are worth 1, mod 3. Then we have (1+1-1=1) for the first bracket, and (1+1+1=0) for the second, i.e. divisibility by 3 in the second bracket
Assume exactly one is worth 0. Then we have (1+0-1=0) for the first bracket, and (1+0+1=2), i.e. also divisibility by 3, this time in the first bracket.
So yes, N is always divisible by 96.
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Posted by broll
on 2016-06-05 12:28:07 |
possible solution
