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Let's first recall some Pythagorean Triples :
$\displaystyle (3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61) $ etc .
Can you see a pattern ? In every triple , we must find one integer from $\displaystyle (x,y,z) $ , such that it is the mutiple of $\displaystyle 5 $ .
$\displaystyle 5,5,15,40,60$ etc .
I would like you to prove this fact , the proof shouldn't be so long , i believe ...
Let (a,b,c) be a pythagorean triple. Observe that $\displaystyle x^2 \equiv 0,1,4$ (mod) 5.
Suppose neither $\displaystyle a,b,c\equiv 0$ (mod) 5. Then $\displaystyle a^2,b^2,c^2\equiv 1,4$ (mod) 5
But then we can never have $\displaystyle a^2+b^2\equiv c^2$ (mod) 5
Contradiction.
The integers $\displaystyle m^2-n^2,\ 2mn,\ m^2+n^2$ form a Pythagorean triple.Quote:
Originally Posted by simplependulum
Consider modulo 5.
If $\displaystyle m\equiv 0\ (mod\ 5)$ or $\displaystyle n\equiv 0\ (mod\ 5)$, then $\displaystyle 2mn\equiv 0\ (mod\ 5)$.
If $\displaystyle m\equiv\pm n\ (mod\ 5)$, then $\displaystyle m^2-n^2\equiv 0\ (mod\ 5)$.
If $\displaystyle m\equiv\pm2\ (mod\ 5)$ and $\displaystyle n\equiv\pm1\ (mod\ 5)$ or $\displaystyle m\equiv\pm1\ (mod\ 5)$ and $\displaystyle n\equiv\pm2\ (mod\ 5)$, then $\displaystyle m^2+n^2\equiv 0\ (mod\ 5)$.
This concludes all the cases.
Hello,
The proof is trivial. Let $\displaystyle x \equiv a \pmod{5}$, $\displaystyle y \equiv b \pmod{5}$, $\displaystyle z \equiv c \pmod{5}$. Now consider a Pythagorean triple :
$\displaystyle x^2 + y^2 = z^2$
Since all terms $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ are modulo $\displaystyle 5$, we may equate and say that :
$\displaystyle a^2 + b^2 \equiv c^2 \pmod{5}$
According to the conjecture, this congruence may only be satisfied if and only if either $\displaystyle a, b, c \equiv 0 \pmod{5}$ (if one of these integers is divisible by $\displaystyle 5$). Recall the quadratic residues modulo $\displaystyle 5$ :
$\displaystyle 0^2 \equiv 0 \pmod{5}$
$\displaystyle 1^2 \equiv 1 \pmod{5}$
$\displaystyle 2^2 \equiv 4 \pmod{5}$
$\displaystyle 3^2 \equiv 4 \pmod{5}$
$\displaystyle 4^2 \equiv 1 \pmod{5}$
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) :
$\displaystyle 0^2 + 0^2 = 0^2 \pmod{5}$
$\displaystyle 0^2 + 1^2 = 1^2 \pmod{5}$
$\displaystyle 0^2 + 1^2 = 4^2 \pmod{5}$
$\displaystyle 0^2 + 2^2 = 2^2 \pmod{5}$
$\displaystyle 0^2 + 2^2 = 3^2 \pmod{5}$
$\displaystyle 0^2 + 3^2 = 2^2 \pmod{5}$
$\displaystyle 0^2 + 3^2 = 3^2 \pmod{5}$
$\displaystyle 0^2 + 4^2 = 1^2 \pmod{5}$
$\displaystyle 0^2 + 4^2 = 4^2 \pmod{5}$
$\displaystyle 1^2 + 0^2 = 1^2 \pmod{5}$
$\displaystyle 1^2 + 0^2 = 4^2 \pmod{5}$
$\displaystyle 1^2 + 2^2 = 0^2 \pmod{5}$
$\displaystyle 1^2 + 3^2 = 0^2 \pmod{5}$
$\displaystyle 2^2 + 0^2 = 2^2 \pmod{5}$
$\displaystyle 2^2 + 0^2 = 3^2 \pmod{5}$
$\displaystyle 2^2 + 4^2 = 0^2 \pmod{5}$
$\displaystyle 3^2 + 0^2 = 2^2 \pmod{5}$
$\displaystyle 3^2 + 0^2 = 3^2 \pmod{5}$
$\displaystyle 3^2 + 1^2 = 0^2 \pmod{5}$
$\displaystyle 3^2 + 4^2 = 0^2 \pmod{5}$
$\displaystyle 4^2 + 0^2 = 1^2 \pmod{5}$
$\displaystyle 4^2 + 0^2 = 4^2 \pmod{5}$
$\displaystyle 4^2 + 2^2 = 0^2 \pmod{5}$
$\displaystyle 4^2 + 3^2 = 0^2 \pmod{5}$
$\displaystyle 2^2 + 1^2 = 0^2 \pmod{5}$
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 $\displaystyle 5$ ... except the first one. However, apart from $\displaystyle 0^2 + 0^2 = 0^2$ which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by $\displaystyle 25$ cannot possibly be divisible by $\displaystyle 25$ (the proof is left as an exercise to the reader (Cool)).
Therefore the conjecture is verified and valid.
$\displaystyle \mathfrak{QED}$
(okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
Any integer is congruent to $\displaystyle 0,\pm 1,\pm 2$ modulo 5 (complete residue system). So any square is congruent modulo 5 to $\displaystyle 0,\pm 1$.
Assume that both $\displaystyle a^2$ and $\displaystyle b^2$ are not congruent modulo 5 to $\displaystyle 0$, then $\displaystyle a^2+b^2\equiv 0,2,$ or $\displaystyle -2(mod\ 5)$.
Since $\displaystyle c^2=a^2+b^2$ and $\displaystyle c^2 $ is a square, we must have $\displaystyle c^2\equiv 0(mod\ 5)$.
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 $\displaystyle x^2 + y^2 = z^2 $ can be represented by $\displaystyle (m^2 - n^2 , 2mn , m^2 + n^2 ) $ . $\displaystyle x $ is odd while $\displaystyle y $ is even , $\displaystyle m , n $ are coprime and one odd one even . (It is true but the proof not included here . )
Then take modulo $\displaystyle 5 $ , we have
$\displaystyle m^2 - n^2 = (m-n)(m+n) $
$\displaystyle m^2 + n^2 \equiv m^2 - 4n^2 = (m-2n)(m+2n) $
consider the product $\displaystyle xyz $ , assume $\displaystyle n $ is prime to $\displaystyle 5 $ , otherwise , $\displaystyle y $ is divisible by $\displaystyle 5 $ .
We have $\displaystyle xyz \equiv 2n(m-2n)(m-n)(m)(m+n)(m+2n) \equiv 2n^5 (s-2)(s-1)(s)(s+1)(s+2) \bmod{5} $ where $\displaystyle s \equiv m(n^{-1}) \bmod{5}$ . We can see that the product $\displaystyle (s-2)(s-1)(s)(s+1)(s+2) $ is divisible by $\displaystyle 5$ so is $\displaystyle xyz $ , this finishes the proof .
