I wrote a sort of 'multigrade' search program (in 2004) to look for things like this. It found two degree 16 examples; I verified the smaller. N = 134810 = 2 5 13 17 61 = square-sum of 343 131, 353 101, 359 77, & 367 11 N = 1200890 = 2 5 29 41 101 = SqSm of 827 719, 953 541, 1049 317, 1087 139 the search to N=10M only found one more, 25*134810. [My search could be improved a lot, by concentrating on products of 4K+1 primes. I simply tried all N = 8K+2 not divisible by 3, 7, or 11; followed by the nested loops to require 4 reps as A^2+B^2; and then some additional checks.] The product of X +- 11,77,101,131,343,353,359,367 is the 16th degree poly (1 -539240 113112960316 -11446255449625880 557362201565582307046 -11360452056407446195762520 95576018314432876073784728956 -275534058738780661111020240872360 31960299178050105687332223346003681) with interspersed 0s. It can be calculated in 4 squarings: (((X^2 - 67405)^2 - 3525798096)^2 - 533470702551552000)^2 - 220156343572527011594632754626560000 factors: 67405 = (5 13 17 61) 3525798096 = (2.4 3.2 17 1049 1373) 533470702551552000 = (2.12 3.4 5.3 17 229 3304237) 469208209191321600 = (2.14 3.6 5.2 7.2 11.2 13 19 29 37) = sqrt(last subtrahend) coverage mod 17: 6 8 1 5 3 4 2 7. ---------- Quoting Warren Smith <warren.wds@gmail.com>:
In the book by Crandall & Pomerance, Phil Carmody is mentioned as having found a lot of polynomials of the form Poly(x) = (((x^2-a)^2-b)^2-c)^2-d with all 16 roots being integers. [Is it possible to get 32 integer roots at the next stage?]
Might Phil inform us more about how he did that and what are these polynomials?
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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