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2018-12-14, 21:17   #4
sweety439

Nov 2016

247910 Posts

Quote:
 Originally Posted by sweety439 Thus, for R4: (we should find n such that all of these formulas take prime values)
e.g. for k=5, 5*4^n-1 has neither trivial factors nor algebraic factors, thus we need n such that 5*4^n-1 takes prime value.

e.g. for k=7, 7*4^n-1 has trivial factor of 3, thus we should take out this factor, and this formula is (7*4^n-1)/3, we need n such that this formula takes prime value. (7*4^n-1 has no algebra factors)

e.g. for k=9, 9*4^n-1 = (3*2^n-1) * (3*2^n+1), thus we need n such that both these two formulas take prime values. (9*4^n-1 has no trivial factors)

e.g. for k=25, 25*4^n-1 has trivial factor of 3, thus we should take out this factor, and this formula is (25*4^n-1)/3, besides, (25*4^n-1)/3 = (5*2^n-1) * (5*2^n+1)/3 (for even n) or (5*2^n-1)/3 * (5*2^n+1) (for odd n), thus we need n such that both these two formulas take prime values.

Last fiddled with by sweety439 on 2018-12-14 at 21:18