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Old 2020-05-18, 13:15   #12
R.D. Silverman
 
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Quote:
Originally Posted by Dr Sardonicus View Post
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.
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Old 2020-05-19, 08:01   #13
kruoli
 
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Originally Posted by JeppeSN View Post
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.
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Old 2020-05-19, 13:53   #14
Dr Sardonicus
 
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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