Number Theory , an easy proof .

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August 1st 2010, 07:53 AM
simplependulum

Number Theory , an easy proof .

Let's first recall some Pythagorean Triples :

$(3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61)$ etc .

Can you see a pattern ? In every triple , we must find one integer from $(x,y,z)$ , such that it is the mutiple of $5$ .

$5,5,15,40,60$ etc .

I would like you to prove this fact , the proof shouldn't be so long , i believe ...
August 1st 2010, 08:10 AM
Dinkydoe

Let (a,b,c) be a pythagorean triple. Observe that $x^2 \equiv 0,1,4$ (mod) 5.

Suppose neither $a,b,c\equiv 0$ (mod) 5. Then $a^2,b^2,c^2\equiv 1,4$ (mod) 5

But then we can never have $a^2+b^2\equiv c^2$ (mod) 5

Contradiction.
August 1st 2010, 08:15 AM
alexmahone

Quote:

Originally Posted by simplependulum

Let's first recall some Pythagorean Triples :

$(3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61)$ etc .

Can you see a pattern ? In every triple , we must find one integer from $(x,y,z)$ , such that it is the mutiple of $5$ .

$5,5,15,40,60$ etc .

I would like you to prove this fact , the proof shouldn't be so long , i believe ...

The integers $m^2-n^2,\ 2mn,\ m^2+n^2$ form a Pythagorean triple.

Consider modulo 5.

If $m\equiv 0\ (mod\ 5)$ or $n\equiv 0\ (mod\ 5)$ , then $2mn\equiv 0\ (mod\ 5)$ .

If $m\equiv\pm n\ (mod\ 5)$ , then $m^2-n^2\equiv 0\ (mod\ 5)$ .

If $m\equiv\pm2\ (mod\ 5)$ and $n\equiv\pm1\ (mod\ 5)$ or $m\equiv\pm1\ (mod\ 5)$ and $n\equiv\pm2\ (mod\ 5)$ , then $m^2+n^2\equiv 0\ (mod\ 5)$ .

This concludes all the cases.
August 1st 2010, 08:29 AM
Bacterius

Hello,
The proof is trivial. Let $x \equiv a \pmod{5}$ , $y \equiv b \pmod{5}$ , $z \equiv c \pmod{5}$ . Now consider a Pythagorean triple :

$x^2 + y^2 = z^2$

Since all terms $a$ , $b$ and $c$ are modulo $5$ , we may equate and say that :

$a^2 + b^2 \equiv c^2 \pmod{5}$

According to the conjecture, this congruence may only be satisfied if and only if either $a, b, c \equiv 0 \pmod{5}$ (if one of these integers is divisible by $5$ ). Recall the quadratic residues modulo $5$ :

$0^2 \equiv 0 \pmod{5}$
$1^2 \equiv 1 \pmod{5}$
$2^2 \equiv 4 \pmod{5}$
$3^2 \equiv 4 \pmod{5}$
$4^2 \equiv 1 \pmod{5}$

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) :

$0^2 + 0^2 = 0^2 \pmod{5}$
$0^2 + 1^2 = 1^2 \pmod{5}$
$0^2 + 1^2 = 4^2 \pmod{5}$
$0^2 + 2^2 = 2^2 \pmod{5}$
$0^2 + 2^2 = 3^2 \pmod{5}$
$0^2 + 3^2 = 2^2 \pmod{5}$
$0^2 + 3^2 = 3^2 \pmod{5}$
$0^2 + 4^2 = 1^2 \pmod{5}$
$0^2 + 4^2 = 4^2 \pmod{5}$
$1^2 + 0^2 = 1^2 \pmod{5}$
$1^2 + 0^2 = 4^2 \pmod{5}$
$1^2 + 2^2 = 0^2 \pmod{5}$
$1^2 + 3^2 = 0^2 \pmod{5}$
$2^2 + 0^2 = 2^2 \pmod{5}$
$2^2 + 0^2 = 3^2 \pmod{5}$
$2^2 + 4^2 = 0^2 \pmod{5}$
$3^2 + 0^2 = 2^2 \pmod{5}$
$3^2 + 0^2 = 3^2 \pmod{5}$
$3^2 + 1^2 = 0^2 \pmod{5}$
$3^2 + 4^2 = 0^2 \pmod{5}$
$4^2 + 0^2 = 1^2 \pmod{5}$
$4^2 + 0^2 = 4^2 \pmod{5}$
$4^2 + 2^2 = 0^2 \pmod{5}$
$4^2 + 3^2 = 0^2 \pmod{5}$
$2^2 + 1^2 = 0^2 \pmod{5}$

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 $5$ ... except the first one. However, apart from $0^2 + 0^2 = 0^2$ which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by $25$ cannot possibly be divisible by $25$ (the proof is left as an exercise to the reader (Cool)).

Therefore the conjecture is verified and valid.

$\mathfrak{QED}$

(okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
August 1st 2010, 11:31 AM
melese

Any integer is congruent to $0,\pm 1,\pm 2$ modulo 5 (complete residue system). So any square is congruent modulo 5 to $0,\pm 1$ .
Assume that both $a^2$ and $b^2$ are not congruent modulo 5 to $0$ , then $a^2+b^2\equiv 0,2,$ or $-2(mod\ 5)$ .
Since $c^2=a^2+b^2$ and $c^2$ is a square, we must have $c^2\equiv 0(mod\ 5)$ .
August 1st 2010, 08:01 PM
simplependulum

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 $x^2 + y^2 = z^2$ can be represented by $(m^2 - n^2 , 2mn , m^2 + n^2 )$ . $x$ is odd while $y$ is even , $m , n$ are coprime and one odd one even . (It is true but the proof not included here . )

Then take modulo $5$ , we have

$m^2 - n^2 = (m-n)(m+n)$

$m^2 + n^2 \equiv m^2 - 4n^2 = (m-2n)(m+2n)$

consider the product $xyz$ , assume $n$ is prime to $5$ , otherwise , $y$ is divisible by $5$ .

We have $xyz \equiv 2n(m-2n)(m-n)(m)(m+n)(m+2n) \equiv 2n^5 (s-2)(s-1)(s)(s+1)(s+2) \bmod{5}$ where $s \equiv m(n^{-1}) \bmod{5}$ . We can see that the product $(s-2)(s-1)(s)(s+1)(s+2)$ is divisible by $5$ so is $xyz$ , this finishes the proof .