Three months ago, in the June riddle, I said that
the riddle was inspired in part by a suggestion from James Ge. In fact, James's
original question to me related to the naturals, not to finite fields. The
question therefore becomes "Given a natural, n, how many Pythagorean
triplets does n participate in?".
This question can be divided into two: - How many (ordered) pairs of positive integers, (
*a*,*b*), are there, such that*a*^{2}+*n*^{2}=*b*^{2}? - How many (unordered) pairs of positive integers, {
*a*,*b*}, are there, such that*a*^{2}+*b*^{2}=*n*^{2}?
n^{2} with
an arbitrary natural, m, and ask:
- How many (ordered) pairs of positive integers, (
*a*,*b*), are there, such that*a*^{2}+*m*=*b*^{2}? - How many (unordered) pairs of positive integers, {
*a*,*b*}, are there, such that*a*^{2}+*b*^{2}=*m*?
m is fully known.
The problem with asking these two questions as monthly riddles is that the former is too easy and the latter is too difficult. The reader is welcomed to answer them as bonus questions. As a monthly riddle, instead of these two questions, I decided to ask something slightly different, namely to prove a lemma that is a major stepping stone in proving the second question. The lemma is:
Prove that for any natural *x*and*y*are co-prime.*i**x*=*y*(mod*n*).*x*^{2}+*y*^{2}=*n*.
As always, we ask for original solutions only, not ones based on prior familiarity or Internet searching. |
## List of solvers:Pei Wu (27 September 01:10) |

Elegant and original solutions can be submitted to
the puzzlemaster at __riddlesbrand.scso.com__.
Names of solvers will be posted on this page. Notify if you don't want
your name to be mentioned.

The solution will be published at the end of the month.

Enjoy!