Question

Is there a pythagorean triplet (a^2+b^2=c^2) such that a and b are odd, and c is even?

Answer 1

Well, one of a and b is odd. C is always odd. So, no.

Answer 2

http://en.wikipedia.org/wiki/Pythagorean_triple You can even check it here.

Answer 3

What do you mean C is always odd?

Answer 4

yes. C is always an odd number.

Answer 5

Lol @quakeash is lying

Answer 6

Well, look at it this way. If c were even, then c^2 is even. This implies that both a^2 and b^2 are even. This in turn implies that a and b are also even. Thus, you can't have an odd a, b and an even c.

Answer 7

if a was odd, a^2 would be odd
if b was odd, b^2 would also be odd

odd + odd = even?

Answer 8

That is true, but remember, if c were even, this eventually implies that a, and b are also even. Let's look at a proof by contradiction. Suppose c were even, so that 4 must divide a^2+b^2. Now, if both a and b are odd, we can show (using some number theory), that 4 can not divide a^2+b^2. This is a contradiction, so this shows that if a, b, are odd, then c can not be even.

Answer 9

Thanks, could you perhaps go into this number theory? :D

Answer 10

Do you know modular arithmetic yet?

Answer 11

yeah

Answer 12

Basically, suppose a, b are odd. This means that they are either \(1\pmod4\) or \(3\pmod4\). If we square these, we find that both \(1^2\equiv3^2\equiv1\pmod4\). So \(a^2+b^2\equiv 1+1\equiv2\pmod4\). This isn't divisible by 4, and leads us to our contradiction.

Answer 13

Great, thanks!

Answer 14

You're welcome.

This also leads to showing that one of a, b is even (for primitive pythagorean triplets). For if they were both odd, then c^2 must be even. But we just showed that can't happen! If both were even, it can't be a primitive triple! Thus, one of them must be even, and the other odd.

Answer 15

Thanks for the great help! I'm using this to prove that for every pythagorean triplet, the incircle radius will be an integer, just needed to confirm this case wasn't possible.

Answer 16

This is just another way of saying that every right triangle has integer radius.

Anyways it can be shown (very easily) that every pyth triplet of the form \((a,b,c)\) can be written as:

\(a= m^2-n^2\)
\(b=2mn\)
\(c= m^2+n^2\)

where \( m,n \in \mathbb{N} \) and \(m>n\)

What you want clearly follows from this :)

Answer 17

How can this be shown?

Answer 18

If \((a, b,c)\) is the triple then \(x^2+y^2=1\) where \(a/c=x\) and \(b/c=y \)

So, \(y^2=(1+x)(1-x) \implies \frac y{(1+x)} = \frac{(1-x)} y=t (say)\)

Where \(t =\frac m n\) ( a rational)

This will give you two simultaneous equation,

\(tx-y=-t\) and \( x+ty=1 \)

Sove them and substitute for x,y and t you will get
\[ \frac a c = \frac {m^2-n^2}{m^2+n^2},\quad\quad\quad \frac b c = \frac {2mn}{m^2+n^2} \]

and this follows what I said before.