# Thread: Pythagorean Triples!

1. ## Pythagorean Triples!

Dear all,
i have recently noticed that all Pythagorean Triples include ONE and only one number that is a multiple of 5. but i have no proof for it.

could anybody help me on this,
thank you.

2. ## Re: Pythagorean Triples!

Hey zenith20.

The formal way of stating this would be to prove that for:

a^2 + b^2 = c^2 then a = 5x (x is an integer) and gcd(a,b) = gcd(a,c) = 1 since 5 is a prime number.

3. ## Re: Pythagorean Triples!

$\displaystyle (15,20,25)$ is a pythogorean triple with more than one multiple of 5. Hence I will assume you are talking about primitive pythogorean triples.

We will prove your claim by contradiction. By definition, the primitive triples should not share a common factor. Suppose two of them shared a common factor of 5, the Pythogorus theorem would then imply that the third one should also contain a factor of 5. Then all of them would have a common factor of 5 contradicting primitiveness of the triple.

4. ## Re: Pythagorean Triples!

many thanks to Isomorphism and Chiro,
now i know that there could not be 'more than One'. but still i can not prove why 'there IS one'. i mean we can not have primitive Pythagorean triples 'without a multiple of 5' number.

5. ## Re: Pythagorean Triples!

Originally Posted by zenith20
many thanks to Isomorphism and Chiro,
now i know that there could not be 'more than One'. but still i can not prove why 'there IS one'. i mean we can not have primitive Pythagorean triples 'without a multiple of 5' number.
That can be answered by simple modular arithmetic:

Consider a pythogorean triple such that $\displaystyle x^2 + y^2 = z^2$. Let us supposed that neither x nor y has a factor of 5. Then observe that $\displaystyle x^2, y^2$ can leave a remainder of 1 or 4 when divided by 5.

If $\displaystyle x^2= y^2=1 \mod 5$, then $\displaystyle z^2 = 2 \mod 5$. But a square of a number cannot be 2 mod 5.
If $\displaystyle x^2= y^2=4 \mod 5$, then $\displaystyle z^2 = 3 \mod 5$. But a square of a number cannot be 3 mod 5.

Thus one of the $\displaystyle x^2, y^2$ must be 1 mod 5 and the other one must be 4 mod 5. Adding we get that z is divisible by 5.

So we have proved that if x and y both are not divisible by 5 then z is divisible by 5.

6. ## Re: Pythagorean Triples!

Originally Posted by Isomorphism
That can be answered by simple modular arithmetic:

...So we have proved that if x and y both are not divisible by 5 then z is divisible by 5.
Very nice. This same appoach can be used to prove that for Pyrthagorian primitives either x or y must be divisible by 3, and furthermore that either x or y must be even and z must be odd (using modular 4 arithmetic).