# Thread: Number Theory , an easy proof .

1. ## Number Theory , an easy proof .

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 ...

2. 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

3. Originally Posted by simplependulum
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 ...
The integers $\displaystyle m^2-n^2,\ 2mn,\ m^2+n^2$ form a Pythagorean triple.

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.

4. 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 ).

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)

5. 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 $\displaystyle c^2\equiv 0(mod\ 5)$.

6. 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 .