Is there any truth to this?

Discussion in 'Number Theory' started by PixelEight, Apr 18, 2021.

  1. PixelEight

    PixelEight

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    Say we want to factorize 87 .. we know that 87=3*29
    We know that there exists two perfect squares that are separated by 87 units. These two squares are 169 and 256.
    256-169=87. And if we take the difference of square roots of the two perfect squares (and addition) we get the prime factors
    of 87. eg. sqrt(256)-sqrt(169)=3 ... sqrt(256)+sqrt(169)=29

    So the problem remains of finding these two perfect squares.

    13 21 23 25 27 29 31
    87 100 121 144 169 196 225 256
    34 57 82 109 138 169

    5^2+9 6^2+21 7^2+33 8^2+45 9^2+57 10^2+69

    12n+9+(n+5)^2 = n^2+22n+34

    (11+2n)((floor(sqrt((9+12n)mod(11+2n)))+1))+(floor(sqrt((9+12n)mod(11+2n)))+1)*((floor(sqrt((9+12n)mod(11+2n)))+1)-1)=9+12n

    According to WolframAlpha: n=5

    so: 5^2+22*5+34=169

    sqrt(169+87) - sqrt(169) = 3
    sqrt(169+87) + sqrt(169) = 29
     
    PixelEight, Apr 18, 2021
    #1
  2. PixelEight

    PixelEight

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    Repost:

    Integer Factorization [email protected]
    ===============

    Say we want to factorize 87 .. 87=3*29
    We know that there exists two perfect squares that are separated by 87 units. These two squares are 169 and 256.
    256-169=87. And if we take the difference of square roots of the two perfect squares (and addition) we get the prime factors
    of 87. eg. sqrt(256)-sqrt(169)=3 ... sqrt(256)+sqrt(169)=29

    So the problem remains of finding these two perfect squares.

    13 21 23 25 27 29 31
    87 100 121 144 169 196 225 256
    34 57 82 109 138 169

    5^2+9 6^2+21 7^2+33 8^2+45 9^2+57 10^2+69

    12n+9+(n+5)^2 = n^2+22n+34


    (2n+11)*Quotient[(12n+9),(2n+11)]+Quotient[(12n+9),(2n+11)]*(Quotient[(12n+9),(2n+11)]-1)=12n+9

    According to WolframAlpha: n=5

    so: 5^2+22*5+34=169

    sqrt(169+87) - sqrt(169) = 3
    sqrt(169+87) + sqrt(169) = 29
     
    PixelEight, Jun 8, 2021
    #2
  3. PixelEight

    paulejking

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    I see a lot of math here, but you never stated the overall problem.... something, something, prime factors which have a property, something something. Not really good enough.

    "So the problem remains of finding these two perfect squares."​

    "These two"? You then proceed with a big list of numbers (more than the two you claim) with a ton of perfect squares, followed by a bunch of steps you didn't explain and don't seem to have a rhyme or reason. Then I see the numbers 3 and 29 on the last step. So what? What is your point?
     
    paulejking, Jul 26, 2021
    #3
  4. PixelEight

    Country Boy

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    Yes, 256 and 169 are "perfect squares"- 256= 16^2 and 169= 13^2 so 256- 169= 16^2- 13^2= (16+ 13)(16- 13)= (29)(3). It is also true that 256- 169= 87 so we must have that 87= (29)(3). But could more easily be seen by 87/3= 29.

    What was your purpose posting this?
     
    Country Boy, Jan 16, 2022
    #4
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