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2020-05-18, 13:15   #12
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Dr Sardonicus According to the references in the link, the Diophantine equation 2n - 7 = x2 is called Ramanujan's square equation. He posed the question of whether it has any solutions for n > 15 in 1913. So I guess you're asking, did he understand the difference between math and numerology?
Yes. Ramanujan. But I saw no math related to this equation in the discussion.
All I saw was a lot of blind computation and sequences of numbers. Indeed, the
first use of the word "equation" came in your post.

2020-05-19, 08:01   #13
kruoli

"Oliver"
Sep 2017
Porta Westfalica, DE

251 Posts

Quote:
 Originally Posted by JeppeSN The only pending is 2^74207282 - 3.
For the record, it has no factors smaller than $$2^{32}$$ and no factor found with P-1 using B1=100,000 and B2=5,000,000. I'll be running some ECM curves on it since it should be more efficient than on mersenne numbers.

2020-05-19, 13:53   #14
Dr Sardonicus

Feb 2017
Nowhere

22×72×17 Posts

Quote:
 Originally Posted by R.D. Silverman Yes. Ramanujan. But I saw no math related to this equation in the discussion. All I saw was a lot of blind computation and sequences of numbers. Indeed, the first use of the word "equation" came in your post.
Here's some basic math regarding the equation 2n - 7 = x2. Rewriting as

x2 - 2n = -7

we have two cases:

x2 - y2 = -7 if n is even, and

x2 - 2*y2 = -7 if n is odd.

In both cases we want solutions where y is a power of 2.

In the first case, there is only one solution in positive integers x and y, x = 3 and y = 4.

In the second case, there are two sequences of solutions. Again, we want y to be a power of 2. The y-sequences may be described as

xn + yn*t = lift((2*t +/- 1)*Mod(3 + 2*t,t2 - 2)n), n = 0, 1, 2, ...

The y-sequences are

y0 = 2, y1 = 8, yk+2 = 6*yk+1 - yk (next terms 46, 268, 1562,...)

and

y0 = 2, y1 = 4, yk+2 = 6*yk+1 - yk (next terms 22, 128, 746,...)

Because of the factors 2*t +/- 1 in the explicit formula, the resulting sequences don't have the nice divisibility properties of the coefficients from the powers of a unit, so ruling out powers of 2 is correspondingly more difficult.

 2020-06-15, 15:26 #15 JeppeSN     "Jeppe" Jan 2016 Denmark 2·71 Posts This may seem silly, but I ran a test with PFGW (slow, one CPU thread) on the remaining number. It took several weeks. Here is the result (PFGW does a 3-PRP test by default): Code: PFGW Version 4.0.1.64BIT.20191203.Win_Dev [GWNUM 29.8] Resuming at bit 72193575 2^74207282-3 is composite: RES64: [B486B8C605802535] (60360.4698s+0.0144s) (The 60k seconds is only for the part between iteration ~72M to ~74M.) So now the residue is here, if someone some day should want to repeat this long calculation. /JeppeSN
2020-06-15, 17:13   #16
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by JeppeSN This may seem silly,
We agree

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