NeilB & I just reduced this to a special case of x^(1/k) == x (x - 2) <=> Log[x - 2, x] == k/(1 - k), which follows from elementary algebra. --rwg On Sat, Nov 2, 2013 at 4:41 PM, Bill Gosper <billgosper@gmail.com> wrote:

NeilB ran an interesting experiment on the OEIS. One of his lesser results:

In[642] := Solve[x == Sqrt[1 + Sqrt[1 + Sqrt[Sqrt[Sqrt[x]]]]]] Out[642] = {{x -> Root[1 + #1 + #1^2 + #1^3 + #1^4 + #1^5 + #1^6 + #1^7 + #1^8 + #1^9 + #1^10 + #1^11 + #1^12 + #1^13 + #1^14 - 255 #1^15 - 255 #1^16 + 769 #1^17 + 769 #1^18 - 1023 #1^19 - 1023 #1^20 + 769 #1^21 + 769 #1^22 - 351 #1^23 - 351 #1^24 + 97 #1^25 + 97 #1^26 - 15 #1^27 - 15 #1^28 + #1^29 + #1^30 &, 2]}}

In[659]:= Log[x^2 - 2, x] == -8/15 /. %642[[1]]

Out[659]= Log[Root[1 + #1 + #1^2 + #1^3 + #1^4 + #1^5 + #1^6 + #1^7 + #1^8 + \ #1^9 + #1^10 + #1^11 + #1^12 + #1^13 + #1^14 - 255 #1^15 - 255 #1^16 + 769 #1^17 + 769 #1^18 - 1023 #1^19 - 1023 #1^20 + 769 #1^21 + 769 #1^22 - 351 #1^23 - 351 #1^24 + 97 #1^25 + 97 #1^26 - 15 #1^27 - 15 #1^28 + #1^29 + #1^30 &, 2]]/ Log[-2 + Root[ 1 + #1 + #1^2 + #1^3 + #1^4 + #1^5 + #1^6 + #1^7 + #1^8 + #1^9 + \ #1^10 + #1^11 + #1^12 + #1^13 + #1^14 - 255 #1^15 - 255 #1^16 + 769 #1^17 + 769 #1^18 - 1023 #1^19 - 1023 #1^20 + 769 #1^21 + 769 #1^22 - 351 #1^23 - 351 #1^24 + 97 #1^25 + 97 #1^26 - 15 #1^27 - 15 #1^28 + #1^29 + #1^30 &, 2]^2] == -(8/15)

In[660]:= N[%, 1111]

Out[660]= True

Whereas this is straightforward to ver[if]y, I don't see how a simplifier could do the LHS except by numerically guessing the -8/15 . Could it somehow Factor those two "logands" into the -8th and 15th powers of a certain root of -1 + 2 x^16 + x^31? --rwg