Triangles (1 Viewer)

[Estel]

The quick question:
Is it possible to have a right-angled triangle where the hypotenuse is not divisible by 5 if all sides have lengths of integer values? Can't think of a counter example right now...

The long question:
Does anyone have a proof for this (if there is no counterexample)?

 

[coca cola]

counter example 21, and 20 for non-hyp sides.

hyp = 29.

umm im let me have a think about a general proof.

 

[coca cola]

well couldn't you go by using the greek's pythagorian triple: (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2, which can be prooved to be true for all positive intergers by using induction.

then the hyp is (m^2 + 1). this is definately NOT divisible by 5 for all intergers.

i.e. m^2 + 1 can not be expressed as 5n in general.

edit: hey have you got a different proof to this?

 

[Estel]

No proof...
the actual question is prove that in a right triangle with sides a, b, and c with gcd(a,b,c)=1 only one of a, b and c will be divisible by 5.

 

[CM_Tutor]

No proof...
the actual question is prove that in a right triangle with sides a, b, and c with gcd(a,b,c)=1 only one of a, b and c will be divisible by 5.

It's easy to show that no more than one of a, b and c will be divisible by 5. If two of them are divisible by 5, then so is the third, in which case gcd(a, b, c) ≠ 1, which is a contradiction. All 3 divisible by 5 leads trivially to the same contradiction. Thus, no more than one of them is divisible by 5 - but I don't see off hand a way to prove that one of them must be divisible by 5.

 

[ngai]

well if u use that (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2
then a bashy method:
m is integer, so m has remainder of r = 0,1,2,3,4 when divided by 5
r = 0, then 2m divisible by 5
r = 1, then m^2 - 1 is divisible by 5
r = 2, then m^2+1
r = 3, then m^2 + 1
r = 4, then m^2 - 1
and then using cm tutor's stuff
u get that theres one and only one is divisible

 

[mojako]

mm what does "gcd(a, b, c)" mean?

 

[Rorix]

greatest common divisor

but im sure you could have found that on Google or something in less than a minute

 

[mojako]

Didnt try google

BTW Estel, what topic does this question come under?

 

[mojako]

I just don't think this is part of the HSC 3U course...
I think I've seen this somewhere but I don't think it's in year 11 or 12:
(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2
and my memory says that it's not to be memorised when I saw it...

 

[Estel]

What topic does this come under?

Interest...
or...
Geometry if you look at the syllabus long and hard enough

ngai
the problem with that is it ignores triads like 7 24 25
So that is not the solution

 

[mojako]

so, did you find the question from a HSC-related textbooks/papers??
if not I wont worry about that
If yes, I might worry, because I know the probability of it appearing on a HSC exam paper is very low, whether it be 2U, 3U or 4U or general maths ^^

 

[Estel]

Not from HSC-related textbooks/papers...
but where's your curiosity?

Expand your horizons

 

[ngai]

ngai
the problem with that is it ignores triads like 7 24 25
So that is not the solution

grr who gave me that dodgy formula?

anyway, try this instead: (i'm gonna use modular arithmetic...)
a^2 + b^2 = c^2
now, for an integer, say x: (== means congruent)
x == 0 (mod 5) gives x^2 == 0 (mod 5)
x == 1 (mod 5) gives x^2 == 1 (mod 5)
x == 2 (mod 5) gives x^2 == 4 == -1 (mod 5)
x == 3 (mod 5) gives x^2 == 9 == -1 (mod 5)
x == 4 (mod 5) gives x^2 == 16 == 1 (mod 5)
so x^2 == 1, -1, or 0 (mod 5)
and if x^2 == 0, then x == 0

if either of a^2 or b^2 == 0 (mod 5), then we have a or b == 0 (mod 5), so we have one divisible by 5

if neither a^2 nor b^2 == 0 (mod 5):
if a^2 == 1, b^2 == 1: c^2 == 2, impossible
if a^2 == 1, b^2 == -1: c^2 == 0, hence c divisible by 5
if a^2 == -1, b^2 == -1: c^2 == -2, impossible

and for those who dunno mod arithmetic, well most simply: "A == B (mod C)" means that when A is divided by C, there is a remainder of B

 

[Estel]

damn...
I really regret not doing those enrichment books :/

 

[mojako]

Not from HSC-related textbooks/papers...
but where's your curiosity?

Expand your horizons

I mean, as an observer who happened to read your thread, which deals with things I haven't learnt (eg modular arithmetics), I don't really care if I can prove that or not.

 

[pc_wizz]

*shudder* triangles

wtf u doing in 3U forum hemi ... ...

 

[Estel]

mojako that doesn't rule out that it doesn't need modular arithmetic tho...
it's quite possible that it doesn't... just because 5 ppl have seen it and not found a solution...

so ppl, if you see a way, do post

 

[Slidey]

stel: I do too, but all is not entirely lost. There is a set of extremely good books on the subject called "The Art of Problem Solving" volumes 1 and 2. They cover basically everything you need to know up to precalculus. That includes conics, vectors, matrices, number theory, combinatorics, complex numbers and modular arithmetic.

http://www.artofproblemsolving.com/Books/AoPS_B_About.php

Here is the index for books one and two, if you want to see exactly what they cover:
Book1: http://www.artofproblemsolving.com/Books/AoPS_B_V1.php
Book2: http://www.artofproblemsolving.com/Books/AoPS_B_V2.php

 

[Archman]

btw i think the proper formula for triads with gcd of 1 is
(p^2-q^2)^2 + (2pq)^2 = (p^2+q^2)^2

(w00t, first post in the ext1 forum)