Let (a,b,c) be a pythagorean triple. Observe that (mod) 5.
Suppose neither (mod) 5. Then (mod) 5
But then we can never have (mod) 5
Contradiction.
Let's first recall some Pythagorean Triples :
etc .
Can you see a pattern ? In every triple , we must find one integer from , such that it is the mutiple of .
etc .
I would like you to prove this fact , the proof shouldn't be so long , i believe ...
Hello,
The proof is trivial. Let , , . Now consider a Pythagorean triple :
Since all terms , and are modulo , we may equate and say that :
According to the conjecture, this congruence may only be satisfied if and only if either (if one of these integers is divisible by ). Recall the quadratic residues modulo :
Now consider all the possibilities (I omitted those that didn't satisfy the conjecture because there's too much of them, this has been double-checked by computer) :
By analyzing the triples that do satisfy the conjecture, we find that all those triples have one, and only one, integer that is a multiple of ... except the first one. However, apart from which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by cannot possibly be divisible by (the proof is left as an exercise to the reader ).
Therefore the conjecture is verified and valid.
(okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
Thanks , indeed , all the solutions here are not so long . I want to include my proof in this post :
Suppose all the primitive solutions of the equation can be represented by . is odd while is even , are coprime and one odd one even . (It is true but the proof not included here . )
Then take modulo , we have
consider the product , assume is prime to , otherwise , is divisible by .
We have where . We can see that the product is divisible by so is , this finishes the proof .