Can anyone see why this formats oddly?

Questions
, ,
1

Hi everyone

Can anyone see why when a solution varible equals 1 the formating doesn’t print, yet for all other numbers it seems fine. Here is the program in its entirety.

Do[a = Solve[(x^2 + y^2 + z^2)/3 == n^2 && y > 0 && x > 0 && z > 0 && 

    x <= y && y <= z, {x, y, z}, Integers]; a = {x, y, z} /. a; 
 a = Flatten[a]; 
 If[Length[a] > 3, a = Delete[a, {{-1}, {-2}, {-3}}]; 
  a = Riffle[a, n, {4, -1, 4}]; a = Partition[a, 4]; 
  If[Length[a] > 1, 
   Print["There are ", Length[a], " solutions when n = ", n]]; 
  If[Length[a] == 1, 
   Print["There is ", Length[a], " solution when n = ", n]]; Do[Print[
\!\(\*SuperscriptBox[\(a[\([q, 1]\)]\), \("\<2\>"\)]\), " + ", 
\!\(\*SuperscriptBox[\(a[\([q, 2]\)]\), \("\<2\>"\)]\), " + ", 
\!\(\*SuperscriptBox[\(a[\([q, 3]\)]\), \("\<2\>"\)]\) " = ", "3x", 
\!\(\*SuperscriptBox[\(n\), \("\<2\>"\)]\)], {q, 1, Length[a]}]], {n, 
  3, 20}]

As you can see there is no superscript printed when the number is 1.

Thanks in advance

Paul.

2

When Mathematica evauates an expression--any expression--it performs first any appropriate simplifications. Since 1 to any non-zero power (even if it is a string!) is 1, Mathematica first does this simplification.

By the way, you might want to learn a bit about things like TextCell and ExpressionCell (along with Defer) and CellPrint so you can use Mathematica’s powerful formatting abilities rather than using the ancient Print function.

3

Thanks David, I took a look at those suggestions and using a mixture of them I have solved the printing format.

Do[a = Solve[(x^2 + y^2 + z^2)/3 == n^2 && y > 0 && x > 0 && z > 0 && 

    x <= y && y <= z, {x, y, z}, Integers]; a = {x, y, z} /. a; 
 a = Flatten[a]; 
 If[Length[a] > 3, a = Delete[a, {{-1}, {-2}, {-3}}]; 
  a = Riffle[a, n, {4, -1, 4}]; a = Partition[a, 4]; 
  If[Length[a] > 1, 
   Print[TextCell[
     Row[{"There are ", Length[a], " solutions when n = ", n}]]]]; 
  If[Length[a] == 1, 
   Print[TextCell[
     Row[{"There is ", Length[a], " solution when n = ", n}]]]]; 
  Do[Print[TextCell[
     Row[{ExpressionCell[a[[q, 1]]]^2, " + ", 
       ExpressionCell[a[[q, 2]]]^2, " + ", 
       ExpressionCell[a[[q, 3]]]^2, " = 3x", 
       ExpressionCell[n]^2}]]], {q, 1, Length[a]}]], {n, 3, 20}]

All is well thank you.

Paul.

4

Great. That’s along the lines of the approach I was thinking of.
--David