Question

[SOLVED] Another fun/interesting problem!

Show that for any Primitive Pythagorean Triple \((a, b, c)\) such that \(a^2+b^2=c^2\), \(5\) divides one of \(a, b,\) or \(c\).

Answer 1

hmm....pythagorean triplets are obtainable like this:\[p ^{2}-q ^{2},2pq,p ^{2}+q ^{2}\]
nut sure how you can prove this from here...

Answer 2

Alternative way to generate triplets: If \(s, t\) are relatively prime, then \[a=st \qquad b={s^2-t^2 \over 2} \qquad c={s^2+t^2 \over 2}\]

Answer 3

that's the same

Answer 4

Ignore what I just said. I read my textbook wrong :(

Answer 5

\[\frac ab\implies\frac s{2t}-\frac t{2s}\]since these are are relativelye prime that means s and t are odd
am I going in the right direction?

Answer 6

Not the way I did it, but it might still work.

Answer 7

therefor\[\{\exists x,y|s=2x+1,t=2y+1\}\]I think...

Answer 8

That's no fun :(

Answer 9

I know... dang linkers

Answer 10

Alternative answers are still encouraged! This question is in my textbook before it teaches about modular arithmetic, and both solution provided by Directrix require the use of modular arithmetic.

Answer 11

Any Pythagorean triple are of the form \(2mn, m^2+n^2, m^2-n^2\) where \( m\gt n\). Now, \(m\) and \( n \) are divisible by \(5\) then \(2mn\) will be and we don't have to do anything.

Otherwise \(m = \pm 1 \pmod 5\) or \( m = \pm 2 \pmod 5 \) and the same thing is true for \(n).


Thus for either of the cases \( m^2+n^2 \) or \(m^2-n^2 \) would be divisible by \(5\).

QED!

The same result holds for for \(3, 4\) too.

Answer 12

EDIT: "Now, m and n are divisible by 5" to "Now, If m or n are divisible by 5"