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Divisibility Patterns in Pythagorean Triples

```Date: 12/08/2003 at 14:15:14
From: Dave
Subject: Pythagorean triples

A student of mine noticed that in every Pythagorean triple (a,b,c such
that a^2 + b^2 = c^2), a or b is divisible by 3, a or b is divisible
by 4, and a, b, or c is divisible by 5 (sometimes the same number
meets more than one of the conditions).

We are trying to determine if this is really true, and if so, how to
prove it.  Could you help?

```

```
Date: 12/08/2003 at 14:57:24
From: Doctor Schwa
Subject: Re: Pythagorean triples

Hi Dave -

What your student noticed is in fact true.  Let's see if we can prove
it.  The idea is to consider the possible remainders when you divide
by 3 (or 4 or 5).

For example, to prove that at least one of a or b is divisible by 3, I
would work like this:

If a and b are not divisible by 3, then a = 3k + 1 or 3k + 2
and you can prove (by squaring them out) that a^2 leaves remainder
1 when you divide by 3:

9k^2 + 6k + 1 and 9k^2 + 12k + 4 both will leave remainder 1,
since 9k^2, 6k, and 12k are automatically divisible by 3.

Similarly b^2 leaves remainder 1, and so c^2 leaves remainder 2
since c^2 = a^2 + b^2.

But that's not possible.  If c is divisible by 3, then c^2 leaves
remainder 0, and if c is not divisible by 3, then c^2 leaves remainder
1.  Remainder 2 can't happen!  Hence at least one of a and b must be
divisible by 3.

Similar proofs work for 4 or 5.  Searching the Dr. Math archives I found

Pythagorean Triples Divisible by 5
http://mathforum.org/library/drmath/view/51578.html

which gives a hint about the divisibility by 5 problem, in case you
get stuck on it.

Enjoy,

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 12/08/2003 at 17:47:27
From: Dave
Subject: Pythagorean triples

Thank you for responding so quickly.  I have done the proof for the
divisibility by 3, but have some problems with the divisibility by
4.  Letting a = 4k+1 and b = 4k+1 works fine but when you set a = 4k+2
and b = 4k+1 you are left with a remainder of 1 for a^2 + b^2 and it's
possible for c^2 to have a remainder of 1 as well.  Some help would be

```

```
Date: 12/08/2003 at 18:52:11
From: Doctor Schwa
Subject: Re: Pythagorean triples

True, the divisibility by 4 is trickier indeed!  Another page from our
archives has one method of proving it, namely using the a = 2uv,
b = u^2 - v^2, c = u^2 + v^2 fact to show that whichever is the even
one of a and b must be divisible by 4 (since u and v are not both odd):

Pythagorean Triples
http://mathforum.org/library/drmath/view/55757.html

But that requires first proving the more difficult formula!  So, how

If all of a,b,c are even, divide all by 2 until you have at least
one odd.  You must then have one of a and b odd--let's call that
one a, one even--call that b, and then c must be odd as well.

So b^2 = c^2 - a^2 = (c - a)(c + a).

Now if c and a are both remainder 1 when we divide by 4, then c - a is
a multiple of 4 and c + a is a multiple of 2, so (c - a)(c + a) is a
multiple of 8.

If c and a are both remainder 3, a similar argument holds.

And if one is remainder 1 and the other remainder 3, then (c + a) is a
multiple of 4, and (c - a) is a multiple of 2.

In any case, (c - a)(c + a) is a multiple of 8.  So if b = 4k+2 =
2*odd, b^2 will be 4*odd^2, which cannot be a multiple of 8.  Thus, b
must be divisible by 4.

I hope that helps clear things up!

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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