Questions for Solution: 9636

9636. (Charles L. Dodgson, M.A.)—If 3 numbers, not in Arithmetical Progression, be such that their sum is a multiple of 3: prove that the sum of their squares is also the sum of another set of 3 squares, the two sets having no common term.

Comment on Solution

Mr. Dodgson states that, in this solution, Prof. Zerr “takes a single special instance of 3 numbers, and seems to think that the theorem, since it is true in this single instance, is thereby proved to b true universally.” He submits the follwowing theorem, and asks whether Professor Zerr would consider the appended proof a sound logical one.

“(Theorem.) If 3 numbers be such that their sum is a multiple of 7, the sum of their squares is a multiple of 9.

“(Proof.) Let m, $2 m$ , $11 m$ be the 3 numbers. Then $m + 2 m + 11 m = 7 \times 2 m$ . Also, $m^{2} + {(2 m)}^{2} + {(11 m)}^{2} = 126 m^{2} = 9 \times 14 m^{2}$ .”

We shall be glad to have a further solution of the Question.