Re: Integers that are the sum of 2 nonzero squares in exactly 2 ways

*To*: mathgroup at smc.vnet.net*Subject*: [mg125946] Re: Integers that are the sum of 2 nonzero squares in exactly 2 ways*From*: Dana DeLouis <dana01 at me.com>*Date*: Sun, 8 Apr 2012 04:18:25 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

> I don't know if this answers anything, but the factorization pattern of such numbers appears to be as follows. > All primes of the form 4n+3 > occur to even powers. Hi. I can't add much, but this article was very interesting... http://mathworld.wolfram.com/SumofSquaresFunction.html One of the codes along the line of 4n+3 was the following: MySquaresRUnordered[n_Integer?NonNegative] := Module[ {b, p, q, primes, powers}, {primes, powers} = Transpose[FactorInteger[n]]; mod1 = Flatten[Position[primes, _?(Mod[#, 4] == 1 &)]]; mod3 = Flatten[Position[primes, _?(Mod[#, 4] == 3 &)]]; If[Or @@ (OddQ /@ powers[[mod3]]), 0, (*Else*) b = Times @@ (1 + # & /@ powers[[mod1]]); If[EvenQ[b], b, b + 1]/2 ] ] But this still considers zero a solution, which the Op does not want. This is not faster, but just interesting. A workaround might be that we want those solutions that equal length 2, but where the square root of the number is not an integer. For example, 400 has two solutions: PowersRepresentations[400,2,2] {{0,20},{12,16}} But... the square root of 400 is the integer 20, so this would not be a valid solution. However, we also need those of length 3, but where the square of the number is an integer. For example, 625 has 3 solutions. but since 25^2 = 625, we can disregard this solution, and assume the remaining 2 solutions are valid. PowersRepresentations[625,2,2] {{0,25},{7,24},{15,20}} Sqrt[625] 25 So, with two aux functions: NotSquare[n_] := Not[IntegerQ[Sqrt[n]]] IsSquare[n_] := IntegerQ[Sqrt[n]] For this idea, you have a table of numbers.... tbl=Table[{n,MySquaresRUnordered[n]},{n,820}]; tt=Cases[tbl,{_,n_/;n==2}][[All,1]] {25,50,65,85,100,125,130,145,169,170,185,200,205,.. ..} But we disregard those that have an integer as a square root. lst1=Select[tt,NotSquare] {50,65,85,125,130,145,170,185,200,205,...} However, the number 625 is missing from this small list. We need to look at the few that have 3 as a solution, and also HAS an integer as a square root. lst2=Select[Cases[tbl,{_,n_/;n==3}][[All,1]],IsSquare] {625} This is ok for small ranges, but slow when the range is large. Only2[n_] := Piecewise[ {{(SquaresR[2, n] - 4)/8 == 2, IntegerQ[Sqrt[n]]}, {(SquaresR[2, n] - 4)/8 == 1, IntegerQ[Sqrt[n/2]]}, {SquaresR[2, n]/8 == 2, True}}] Select[Range[820],Only2] {50,65,85,125,130,145,170,185,200,205,221,...} = = = = = = = = = = = = = = = Interesting subject :>) Dana DeLouis Mac & Math 8 = = = = = = = = = = = = = = = On Mar 31, 4:44 am, d... at wolfram.com wrote: > On Wednesday, March 28, 2012 5:00:45 AM UTC-5, Cisco Lane wrote: > > I've been looking at integers that are the sum of 2 nonzero squares in exactly 2 ways. The smallest example is 50 = 5^2+5^2=7^2+1^2. The first few terms are 50, 65, 85, 125, 130, 145, .... This is given in OEIS ashttps://oeis.org/A025285 > > > If I plot the pairs {1,50},{2,65},{3,85},... I get a more or less straight line with a slope of about 8.85... In other words, eventually, about one in 8.85 integers qualify. > > > I wonder if there is a theoretical value for this approximate number of 8.85...? > > I don't know if this answers anything, but the factorization pattern of such numbers appears to be as follows. All primes of the form 4n+3 occur to even powers. There are one, two or three other factors. They can take the form > > 2^n*p*q where p and q are distinct primes of the form 4n+1, n an arbitrary nonnegative integer > > 2^(2*n+1)*p^2 with n, p as above > > 2^(2*n)*p^(2*k) for n as above and k satisfying some relation I cannot quite figure out other than it has to be at least 2. There may also be further restrictions on n that depend on k. > > Daniel Lichtblau > Wolfram Research Re: Integers that are the sum of 2 nonzero squares in exactly 2 ways