Originally Posted by

**Media_Man** Wonderboy1953: We all appreciate your patience, but I think I speak for the rest of us posters when I say your original question is extremely difficult to follow. I understand what multigrades are; I understand what magic squares are. But I have yet to understand what exactly it is that you are conjecturing. I *thought* you were claiming that no multigrade of level>2 can be "dotted" with another vector, and create another multigrade of level>2. If that is the conjecture you were intending, then the counterexample I found makes it definitively false, since I took a trigrade, dotted it with (16,4,4,1) and made another trigrade out of it. (Sorry, by the way, I cannot enlighten you with my derivation of this result. I took my own advice and let a computer sift through thousands of candidates and that is one that popped up. You can find the code at the end of this post if you are interested.) Since you are continuing to defend your observation, I can only assume that I have misinterpreted your original question.

I know you are relying on hand calculations in making these observations. I would be more than happy to use a computer to search for a counterexample, if it exists. Obviously, this has no relevance for *proving* a conjecture, but is the easiest way to *disprove* it.

Again, **what exactly is it that you are conjecturing?**

int max=50;

for (int c1=-max; c1<max; c1++){

for (int c2=-max; c2<max; c2++){

for (int c3=-max; c3<max; c3++){

for (int c4=-max; c4<max; c4++){

if ((-c1+4*c2+c3-4*c4==0)

&&(-5*c1*c1+56*c2*c2+29*c3*c3-80*c4*c4==0)

&&(-19*c1*c1*c1+604*c2*c2*c2+631*c3*c3*c3-1216*c4*c4*c4==0)

&&!(c1==c2 && c2==c3 && c3==c4))

System.out.println("c1="+c1+", c2="+c2+", c3="+c3+", c4="+c4);

}}}}