[seqfan] Re: Can more of these terms be found?

[israel at math.ubc.ca]

For an example with 2d+1 digits, you want an integer solution of

(y-x)^2 = 10^d x + y

where 0 <= y < 10^d and 10^(d-2) <= x < 10^(d-1).
Now solving the quadratic equation for x, 

x = y + 10^d/2 - 1/2 sqrt(10^(2d) + 4 y (10^d+1))
  = y + 10^d/2 - z/2
where z^2 = 10^(2d) + 4 y (10^d+1), and so
y = (z^2 - 10^(2d))/(4 (10^d+1)).

For y to be an integer, z must be even, and z^2 == 10^(2d) == 1 mod 
(10^d+1). For 0 <= y < 10^d we need 10^(2d) <= z^2 < 5 * 10^(2d) + 4 * 
10^d. But z = 10^d and z = 10^d+2 produce x=0, so those are not allowed In 
particular, if 10^d+1 is prime, the only solutions of z^2 == 1 mod (10^d+1) 
are z == (+/-) 1 mod 10^(d+1), and the only even z in the interval are 10^d 
and 10^d + 2. So we get no solutions for d = 1 or 2,

d = 3: 10^3+1 = 7 * 11 * 13.  Candidates for z are 1156, 1574, 1728
leading to (y,x) = [84, 6], [369, 82], [496, 132].  The one with x in
the interval [100, 1000) is [369, 82] which is your solution.

d = 4: Candidate for z is 12192 leading to your solution.

d = 5: z =  136366, y = 21489, x = 3306
(21489 - 3306)^2 = 330621489

d = 6: z = 1673268, y = 449956, x = 113322 > 10^5

d = 7: z = 15454548, y = 3471076, x = 743802
(3471076 - 743802)^2 = 7438023471076

d = 8: z = 147058824, y = 29065744, x = 5536332
(29065744 - 5536332)^2 = 553633229065744

d = 9: 20 candidates, of which 6 have x in the desired interval:
(135490884 - 14611762)^2 = 14611762135490884
(159622884 - 19605006)^2 = 19605006159622884
(193545216 - 27553312)^2 = 27553312193545216
(294293121 - 56530882)^2 = 56530882294293121
(371816704 - 83263152)^2 = 83263152371816704
(402366864 - 94674556)^2 = 94674556402366864

Cheers,
Robert Israel





On Aug 8 2013, DAN_CYN_J wrote:

>
>
>Hi all seqfans, 
>
>  
>
> So far only 2 terms are known that will always, I believe, be of odd 
> length 
>
>and one integer per discrete length. 
>
> On the first term subtract the 2 high order digits from the 3 low order 
> digits.
>
>Square the results which will give the original integer.   
>
>82369 ----369-82 =287---287^2 =82369 
>
>1201216 ---1216-120=1096---1096^2 =1201216 
>
>This was just posted on sci-math given the larger integer above. 
>
>I found the smaller one (82369) . 
>
>Is a (9),(11),(13)..  digit possible to calculate? 
>
>Dan   
>
> 
>
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