[math-fun] "Aurifeuillean" factorization of Mersenne^2+8
Neil and I recently noticed
(2^(2*k + 1) - 1)^2 + 8 == (2^(2*k + 1) + 2^(k + 2) + 3) * (2^(2*k + 1) - 2^(k + 2) + 3) Old? --rwg
(I never received my copy of this, but I assume most of you did, since I got an offlist response.) On Tue, Dec 17, 2013 at 6:20 PM, Bill Gosper <billgosper@gmail.com> wrote:
Neil[B] and I recently noticed
(2^(2*k + 1) - 1)^2 + 8 == (2^(2*k + 1) + 2^(k + 2) + 3) * (2^(2*k + 1) - 2^(k + 2) + 3) Old? --rwg
This is pretty dull, since Factor[...] can simply do it. Slightly more interesting is In[182]:= Factor[(2^(k + 1) - 3)^2 - 8]
Out[182]= 1 - 3 2^(2 + k) + 2^(2 + 2 k) (Fails) but both the "odd" and "even" subcases work: In[185]:= Factor[(2^(2*k + 1) - 3)^2 - 8] Out[185]= (1 - 2^(2 + k) + 2^(1 + 2 k)) (1 + 2^(2 + k) + 2^(1 + 2 k)) In[186]:= Factor[(2^(2*k) - 3)^2 - 8] Out[186]= (-1 + 2^(2 k) - 2^(1 + k)) (-1 + 2^(2 k) + 2^(1 + k)) --rwg
On Wed, Dec 18, 2013 at 12:07 AM, Bill Gosper <billgosper@gmail.com> wrote:
(I never received my copy of this, but I assume most of you did, since I got an offlist response.)
It finally arrived along with my copy of the 2nd msg, dated 18hr and 13min later,.
On Tue, Dec 17, 2013 at 6:20 PM, Bill Gosper <billgosper@gmail.com> wrote:
Neil[B] and I recently noticed
(2^(2*k + 1) - 1)^2 + 8 ==
(2^(2*k + 1) + 2^(k + 2) + 3) * (2^(2*k + 1) - 2^(k + 2) + 3) Old? --rwg
This is pretty dull, since Factor[...] can simply do it. Slightly more interesting is In[182]:= Factor[(2^(k + 1) - 3)^2 - 8]
Out[182]= 1 - 3 2^(2 + k) + 2^(2 + 2 k) (Fails) but both the "odd" and "even" subcases work:
In[185]:= Factor[(2^(2*k + 1) - 3)^2 - 8]
Out[185]= (1 - 2^(2 + k) + 2^(1 + 2 k)) (1 + 2^(2 + k) + 2^(1 + 2 k))
In[186]:= Factor[(2^(2*k) - 3)^2 - 8]
Out[186]= (-1 + 2^(2 k) - 2^(1 + k)) (-1 + 2^(2 k) + 2^(1 + k)) --rwg
In the even case, we can generalize 2^_ to x^_:
In[208]:= Factor[(x^(2*k) - 3)^2 - 8] Out[208]= (-1 - 2 x^k + x^(2 k)) (-1 + 2 x^k + x^(2 k)) In[210]:= Expand[%] Out[210]= 1 - 6 x^(2 k) + x^(4 k) But In[209]:= Factor[(x^(2*k + 1) - 3)^2 - 8] Out[209]= 1 - 6 x^(1 + 2 k) + x^(2 + 4 k) In fact, In[210]:= PrimeQ[% /. x -> 6 /. k -> 16] Out[210]= True --rwg
Quoting Bill Gosper <billgosper@gmail.com>: <clip>
On Tue, Dec 17, 2013 at 6:20 PM, Bill Gosper <billgosper@gmail.com> wrote:
Neil[B] and I recently noticed
(2^(2*k + 1) - 1)^2 + 8 ==
(2^(2*k + 1) + 2^(k + 2) + 3) * (2^(2*k + 1) - 2^(k + 2) + 3) Old? --rwg
This is pretty dull, since Factor[...] can simply do it. Slightly more interesting is In[182]:= Factor[(2^(k + 1) - 3)^2 - 8]
Out[182]= 1 - 3 2^(2 + k) + 2^(2 + 2 k) (Fails) but both the "odd" and "even" subcases work:
In[185]:= Factor[(2^(2*k + 1) - 3)^2 - 8]
Out[185]= (1 - 2^(2 + k) + 2^(1 + 2 k)) (1 + 2^(2 + k) + 2^(1 + 2 k))
In[186]:= Factor[(2^(2*k) - 3)^2 - 8]
Out[186]= (-1 + 2^(2 k) - 2^(1 + k)) (-1 + 2^(2 k) + 2^(1 + k)) --rwg
In the even case, we can generalize 2^_ to x^_: In[208]:= Factor[(x^(2*k) - 3)^2 - 8]
Out[208]= (-1 - 2 x^k + x^(2 k)) (-1 + 2 x^k + x^(2 k))
In[210]:= Expand[%]
Out[210]= 1 - 6 x^(2 k) + x^(4 k)
But
In[209]:= Factor[(x^(2*k + 1) - 3)^2 - 8]
Out[209]= 1 - 6 x^(1 + 2 k) + x^(2 + 4 k) In fact,
In[210]:= PrimeQ[% /. x -> 6 /. k -> 16]
Out[210]= True --rwg
Out209 can be morphed to Y^4 - 6 Y^2 + 1 = (Y^2 - 1)^2 - 4 Y^2 = (Y^2 - 2 Y - 1) * (Y^2 + 2 Y - 1). This is similar to the canonical Aurif factorizations for 2^(4K+2)+1 and 3^(6K+3)+1, which can be morphed to factorizations of 4 Y^4 + 1 = (2Y^2+1)^2 - 4Y^2 and 27 Y^6 + 1 = (3Y^2+1) (9Y^4 - 3Y^2 + 1), with the final factor splitting as (3Y^2+1)^2 - 9Y^2. I like to think of these as [1 -6 1] = [1 -2 1] - [4 0] and [4 0 1] = [4 4 1] - [4 0] and [9 -3 1] = [9 6 1] - [9 0], followed by making the subtrahends into squares. Rich
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