1,2

If A*10^m+B is an element, then so is (10^m-A)*10^m+B (e.g. 1233 <-> 8833 or 010100 <-> 990100)

Since A^2+B^2 = A*10^m+B can be written as 10^(2*m)+1 = (2*A-10^m)^2 + (2*B-1)^2 the number of solutions with 2*m digits (necessary leading zeros count) can be reduced to finding the ways 10^(2*m)+1 can be written as sum of 2 squares. For the following results, see A002654. Since 10^(2*m)+1 is odd and has no prime factors of the form 4*r+3 the number of ways writing 10^(2*m)+1 as sum of 2 squares is just tau(10^(2*m)+1) (order matters). Since changing the order does not lead to a new solution (2*A-10^m is always the even square and 2*B-1 is always the odd square) and since the trivial 10^(2*m)+1 = (10^m)^2 + 1^2 leads to the ambiguous A = 0 and B = 1 there are only tau(10^(2*m)+1)/2-1 relevant ways. Because of the transformation from A to (10^m-A) every of these possibilities leads to a pair of solutions. So the number of solutions with 2*m digits is tau(10^(2*m)+1)-2, see A064943

IBM Corp., IBM March 2000 Challenge solution

C. Rivera, Problem 104 of the prime puzzle pages with other approaches

8833 = 88^2 + 33^2, 5882353 = 0588^2 + 2353^2.

with (numtheory): for m from 1 to 10 do: for i in sum2sqr(10^(2*m)+1) do: if i[1] > 1 and i[1] < 10^m then if type(i[1], odd) then a := (i[2]+10^m)/2: b := (i[1]+1)/2: else a := (i[1]+10^m)/2: b := (i[2]+1)/2: fi: print("Length =", 2*m, "Solution =", (10^m-a)*10^m+b): print(Length = 2*m, Solution = a*10^m+b): fi: od: od:

Cf. A064943 for the number of solutions, A055616 for the solutions where leading zeros are not allowed, A055617, A055618, A055619 for some infinite subsequences and A002654 for finding the number of ways writing an integer as sum of two squares.

A101311 is another version.

Sequence in context: A032348 A038824 A038813 this_sequence A101311 A055616 A104971

Adjacent sequences: A064939 A064940 A064941 this_sequence A064943 A064944 A064945

nonn,base

Ulrich Schimke (ulrschimke(AT)aol.com)

Edited by N. J. A. Sloane (njas(AT)research.att.com), Jul 31 2007