RCS's proof is (rewritten so I can follow it): If N>=0 then (5N)^2 < (3N-1)^2 + (4N+1)^2 = (5N)^2 + 2N+2 < (5N+1)^2 (5N)^2 < (3N+2)^2 + (4N)^2 = (5N)^2 + 12N+4 = (5N+1)^2 + 2N+3 < (5N+2)^2 Therefore any even number of form 2N+2 or odd number of form 2N+3 if N>50, is representable in the Wilson a^2+b^2-c^2 form with c=floor(sqrt(a^2+b^2)); and this is all numbers except for some finite set of small integers, which are covered by the computer search I'd done (which was large overkill for that). Actually the exceptional set is {0, 1} so if RCS and I split up the work we can probably handle it even without a computer. Brilliantly simple. (I'd actually thought of the same kind of idea but had somehow managed to confuse myself into thinking it was not adequate...) ===== Now to go a bit further, be nice to prove that not only is every number representable but in fact has an infinite set of representations. Which I shall now show. If r and s fixed integers then for all sufficiently large N (5N+12r+12s)^2 < (3N+20r)^2+(4N+15s)^2 = (5N)^2 + 120(r+s)N+(20r)^2+(15s)^2 = (5N+12r+12s)^2 + (16r-9s)^2 < (5N+12r+12s+1)^2 inspiring (5N+12r+12s)^2 < (3N+20r-1)^2+(4N+15s+1)^2 = = (5N+12r+12s)^2 + (16r-9s)^2 + 2N+2+30s-40r = (5N+12r+12s+1)^2 + (16r-9s)^2 + 1+6s-64r-8N So every even number (2N+2+30s-40r) is representable in an infinite set of Wilsonian ways, e.g. 2N+2>=0 is by considering the infinite set of (r,s) with 30s-40r=0. (5N+12r+12s)^2 < (3N+20r+2)^2+(4N+15s)^2 = = (5N+12r+12s)^2 + (16r-9s)^2 + 12N+4+80r = (5N+12r+12s+1)^2 + (16r-9s)^2 + 2N+3+56r-24s = (5N+12r+12s+2)^2 + (16r-9s)^2 + 32r-48s-8N So every odd number (2N+3+56r-24s) is representable in an infinite set of Wilsonian ways, e.g. 2N+3>=0 is by considering the infinite set of (r,s) with 56r=24s and r>=0. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)