Question 1135584

the sum of squares {{{3n}}} procedure finds all{{{ a}}},{{{b}}},{{{c}}} such that {{{n = a^2 + b^2 + c^2}}} 
It works by repeatedly invoking sum of squares on {{{n-i^2}}} where {{{i}}} is incremented after each iteration from {{{1}}} to square root of {{{n/3}}}


Do you want *exactly* {{{3}}} ways, or *at least* {{{3}}} ways? 

Assuming order isn't considered, and assuming you want it {{{exactly}}} in {{{3 }}}ways, here are a few numbers: 

{{{54}}}, {{{66}}}, {{{81}}},{{{ 86}}}, {{{89}}}, {{{99}}},{{{ 101}}}, {{{110}}}, {{{114}}}, {{{126}}}, {{{131}}}, {{{149}}},{{{ 150}}}, {{{162}}}, {{{166}}}, {{{173}}}, {{{174}}},{{{ 179}}}, {{{182}}}, {{{185}}}, {{{186}}}, {{{216}}}, {{{219}}}, {{{221}}}, {{{222}}}, {{{225}}}, {{{227}}}, {{{233}}}, {{{237}}}, ... 

Example: 

{{{182 = 1^2 + 9^2 + 10^2}}} 
{{{182 = 2^2 + 3^2 + 13^2}}} 
{{{182 = 5^2+ 6^2 + 11^2}}}