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science_man_88 · 2017-01-09, 19:05

most of what I can think of I probably already have posted somewhere like the trivial results that $2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m}}$

edit: yep so trivial I messed up it's statement. basically I was just using the fact that:
1(2x+1)^2 = 4x^2+4x+1 = x(4x+3)+(x+1)
2(2x+1)^2 = 8x^2+8x+2 = x(8x+7)+(x+2)

which technically is:

$2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m+1}}$ because m=0 for the 1 case. so this is just a specific case of a more general algebraic fact.

2017-01-09, 19:05	#1
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	what should I post ? most of what I can think of I probably already have posted somewhere like the trivial results that $2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m}}$ edit: yep so trivial I messed up it's statement. basically I was just using the fact that: 1(2x+1)^2 = 4x^2+4x+1 = x(4x+3)+(x+1) 2(2x+1)^2 = 8x^2+8x+2 = x(8x+7)+(x+2) which technically is: $2^m{M_n}^2 \equiv M_{n-1}+2^m \pmod {M_{n+m+1}}$ because m=0 for the 1 case. so this is just a specific case of a more general algebraic fact. Last fiddled with by science_man_88 on 2018-01-20 at 23:56 Reason: Fixed basic algebra mistakes